Нед. 7–9. Прямые в пространстве, плоскости в пространстве, системы линейных уравнений

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1. Краткое содержание

1.1 Уравнения прямой в пространстве \(\mathbb{R}^3\)

Прямую в пространстве можно однозначно задать точкой на прямой и направлением (ненулевым направляющим вектором).

1.1.1 Векторная и параметрическая формы

Векторное уравнение прямой (vector form) задаёт все точки прямой через точку и направляющий вектор.

• Пусть \(P_0 = (x_0, y_0, z_0)\) — известная точка на прямой.
• Пусть \(\vec{u} = (a, b, c)\) — направляющий вектор (direction vector): ненулевой вектор, параллельный прямой.
• Произвольная точка \(P = (x, y, z)\) на прямой достигается радиусом-вектором \(\vec{OP} = \vec{OP_0} + t\vec{u}\), где \(t\in\mathbb{R}\).

Отсюда параметрические уравнения (parametric form): \[ \begin{cases} x = x_0 + at \\ y = y_0 + bt \\ z = z_0 + ct \end{cases} \] Здесь \(t\) — параметр; при изменении \(t\) точка пробегает всю прямую.

1.1.2 Канонические (симметрические) уравнения

Если \(a,b,c\neq 0\), из параметрических уравнений можно исключить \(t\) и получить канонический вид (symmetric equations): \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \] По нему удобно читать направление \((a,b,c)\) и точку \((x_0,y_0,z_0)\). Если, например, \(a=0\), прямая лежит в плоскости \(x=x_0\), и запись корректируется: \(x=x_0,\ \frac{y-y_0}{b}=\frac{z-z_0}{c}\).

1.2 Взаимное расположение двух прямых

В \(\mathbb{R}^3\) возможны четыре типа:

1. Параллельны (parallel): направления коллинеарны, общих точек нет.
2. Пересекаются (intersecting): не параллельны и имеют ровно одну общую точку.
3. Скрещиваются (skew lines): не параллельны и не пересекаются (лежат в разных плоскостях).
4. Совпадают (coincident): одна и та же прямая.

1.3 Расстояние от точки до прямой

Кратчайшее расстояние от точки до прямой — длина перпендикуляра.

• \(P\) — точка вне прямой;
• прямая задана точкой \(P_0\) и направлением \(\vec{u}\);
• площадь параллелограмма на векторах \(\vec{P_0P}\) и \(\vec{u}\) равна \(||\vec{P_0P}\times \vec{u}||\), а «высота» относительно основания \(||\vec{u}||\) даёт расстояние.

Формула: \[ D = \frac{||\vec{P_0 P} \times \vec{u}||}{||\vec{u}||} \]

1.4 Уравнения плоскости в \(\mathbb{R}^3\)

Плоскость можно задать точкой в плоскости и нормалью — вектором, перпендикулярным плоскости.

1.4.1 Вид «точка + нормаль» (point-normal form)

Плоскость однозначно задаётся точкой \(P_0=(x_0,y_0,z_0)\) и нормалью (normal vector) \(\vec{n}=(a,b,c)\).

Для любой точки \(P=(x,y,z)\) плоскости вектор \(\vec{P_0P}\) ортогонален \(\vec{n}\), поэтому \(\vec{n}\cdot \vec{P_0P}=0\), то есть \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

1.4.2 Общий вид (standard / general form)

После раскрытия скобок получают \[ ax + by + cz = d \] где \((a,b,c)\) — компоненты нормали, а \(d = ax_0 + by_0 + cz_0\).

1.5 Расстояние от точки до плоскости

Расстояние от \(P_1=(x_1,y_1,z_1)\) до плоскости \(ax+by+cz+D=0\): \[ \text{Distance} = \frac{|ax_1 + by_1 + cz_1 + D|}{\sqrt{a^2 + b^2 + c^2}} \] (в числителе — модуль подстановки точки в левую часть уравнения плоскости.)

1.6 Угол между двумя плоскостями

Угол между плоскостями берут как острый угол между их нормалями \(\vec{n_1},\vec{n_2}\):

\[ \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| ||\vec{n_2}||} \]

1.7 Системы уравнений и геометрия

Системы линейных уравнений естественно интерпретировать через пересечения прямых и плоскостей.

1.7.1 Прямые на плоскости (2D)

Система \[ \begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases} \] — это две прямые на плоскости; решения — точки пересечения.

• Единственное решение: прямые пересекаются в одной точке.
• Нет решений: параллельны и различны.
• Бесконечно много решений: совпадают.

1.7.2 Плоскости в пространстве (3D)

Три уравнения с тремя неизвестными — три плоскости; решение — их общее пересечение.

• Единственное решение: общая точка.
• Бесконечно много решений: общая прямая (пучок решений).
• Нет решений: общей точки нет (в том числе «призма» из трёх плоскостей).

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2. Определения

• Направляющий вектор (direction vector): ненулевой \(\vec{u}=(a,b,c)\), параллельный прямой.
• Нормаль (normal vector): вектор \(\vec{n}=(a,b,c)\), перпендикулярный плоскости.
• Параметрические уравнения прямой: \(x=x_0+at,\ y=y_0+bt,\ z=z_0+ct\).
• Канонические уравнения прямой: \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).
• Скрещивающиеся прямые (skew lines): не параллельны и не пересекаются.
• Вид «точка + нормаль» (point-normal form): \(a(x-x_0)+b(y-y_0)+c(z-z_0)=0\).
• Общий вид плоскости (standard form): \(ax+by+cz=d\).

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3. Формулы

• Параметрические уравнения прямой: \[\begin{cases} x = x_0 + at \\ y = y_0 + bt \\ z = z_0 + ct \end{cases}\]
• Канонические уравнения прямой: \[\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\]
• Плоскость: точка + нормаль: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\)
• Общий вид плоскости: \(ax + by + cz + D = 0\)
• Нормаль через векторное произведение: \(\vec{n} = \vec{u} \times \vec{v}\)
• Скалярное произведение (dot product): \(\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\)
• Векторное произведение (cross product): \[\vec{a} \times \vec{b} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix}\]
• Угол между векторами: \[\cos \theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| ||\vec{b}||}\]
• Угол между плоскостями: \[\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| ||\vec{n_2}||}\]
• Расстояние от точки до прямой: \[D = \frac{||\vec{P_0 P} \times \vec{u}||}{||\vec{u}||}\]
• Расстояние от точки до плоскости: \[\text{Distance} = \frac{|ax_1 + by_1 + cz_1 + D|}{\sqrt{a^2 + b^2 + c^2}}\]
• Расстояние между скрещивающимися прямыми: \[d = \frac{|\vec{P_1P_2} \cdot (\vec{v_1} \times \vec{v_2})|}{||\vec{v_1} \times \vec{v_2}||} = \frac{|\vec{P_1P_2} \cdot \vec{n}|}{||\vec{n}||}\]
• Смешанное произведение (объём параллелепипеда): \(V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = |(\vec{a} \times \vec{b}) \cdot \vec{c}|\)
• Площадь треугольника в 3D: \(\text{Area} = \frac{1}{2}||\vec{AB} \times \vec{AC}|| = \frac{1}{2}||\vec{u} \times \vec{v}||\)
• Площадь параллелограмма через \(\det(A^TA)\): \(\text{Area} = \sqrt{\det(A^T A)}\)
• Проекция вектора (vector projection): \[\text{proj}_{\vec{u}}(\vec{v}) = \frac{\vec{v} \cdot \vec{u}}{||\vec{u}||^2} \vec{u}\]
• Отражение вектора (vector reflection): \(\text{ref}_{\vec{n}}(\vec{v}) = 2 \cdot \text{proj}_{\vec{n}}(\vec{v}) - \vec{v}\)
• Тождество двойного векторного произведения: \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\)
• Линейная независимость трёх векторов (определитель): \(\det([\vec{v_1} \ \vec{v_2} \ \vec{v_3}]) \neq 0\)
• Нормирование (unit vector): \[\hat{u} = \frac{\vec{v}}{||\vec{v}||}\]
• Уравнение прямой в отрезках (2D): \(\frac{x}{a} + \frac{y}{b} = 1\)
• Середина отрезка (2D): \(M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
• Середина отрезка (3D): \[ M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right) \]
• Угловой коэффициент (2D): \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
• Угловой коэффициент из угла: \(m = \tan(\theta)\)
• Уравнение прямой через точку с угловым коэффициентом (2D): \(y - y_1 = m(x - x_1)\)
• Перпендикулярные прямые (2D): \(m_2 = -\frac{1}{m_1}\)
• Центроид треугольника: \(C = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
• Формула деления отрезка: \(R = \left(\frac{kx_2+x_1}{k+1}, \frac{ky_2+y_1}{k+1}, \frac{kz_2+z_1}{k+1}\right)\)
• Обратная матрица через присоединённую: \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\)
• Вырожденная матрица: \(\det(A) = 0\)
• Определитель произведения: \(\det(AB) = \det(A)\det(B)\)
• Правило Крамера: \(x_i = \frac{\det(A_i)}{\det(A)}\)
• Перпендикулярность: \(\vec{u} \cdot \vec{v} = 0\)
• Угол между прямой и плоскостью: \[ \sin(\theta) = \frac{|\vec{v} \cdot \vec{n}|}{||\vec{v}|| ||\vec{n}||} \]
• Единственность решения / невырожденность: \(\det(A) \neq 0\)
• Нетривиальные решения однородной системы: \(\det(A) = 0\)
• Модуль векторного произведения (геометрия): \(||\vec{u} \times \vec{v}|| = ||\vec{u}|| ||\vec{v}|| \sin(\theta)\)
• Скалярное произведение (геометрия): \(\vec{u} \cdot \vec{v} = ||\vec{u}|| ||\vec{v}|| \cos(\theta)\)
• Квадрат нормы разности: \(||\vec{a} - \vec{b}||^2 = ||\vec{a}||^2 - 2(\vec{a}\cdot\vec{b}) + ||\vec{b}||^2\)
• Расстояние от точки до прямой (3D): \[ d = \frac{||\vec{AP} \times \vec{v}||}{||\vec{v}||} = \frac{||\vec{P_0 P} \times \vec{u}||}{||\vec{u}||} \]

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4. Примеры

4.1. Уравнение прямой: середина отрезка между осями (Лаба 6, Задание 1)

Find the equation of the straight line, the portion of which between the axes is bisected at the point \((2, -5)\).

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1. Use the Intercept Form: Let the equation of the line be \(\frac{x}{a} + \frac{y}{b} = 1\), where the line intersects the x-axis at \((a, 0)\) and the y-axis at \((0, b)\).
2. Use the Midpoint Formula: The midpoint of the segment between the intercepts is given by \((\frac{a+0}{2}, \frac{0+b}{2}) = (\frac{a}{2}, \frac{b}{2})\).
3. Find the Intercepts: We are given that the midpoint is \((2, -5)\).
   • \(\frac{a}{2} = 2 \implies a = 4\).
   • \(\frac{b}{2} = -5 \implies b = -10\).
4. Write the Equation: Substitute the values of \(a\) and \(b\) into the intercept form.
   • \(\frac{x}{4} + \frac{y}{-10} = 1\).
5. Simplify to General Form: Multiply the entire equation by 20 to clear the denominators.
   • \(5x - 2y = 20\).
   • \(5x - 2y - 20 = 0\).

Ответ: The equation of the line is \(5x - 2y - 20 = 0\).

4.2. Уравнение прямой через пересечение двух прямых и угол с осью \(Ox\) (Лаба 6, Задание 2)

Find the equation of the straight line passing through the intersection of the lines \(3x - y = 5\) and \(2x + 3y = 7\) and making an angle of \(45^\circ\) with the positive direction of the x-axis.

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1. Find the Point of Intersection: Solve the system of two linear equations.
   • From \(3x - y = 5\), we get \(y = 3x - 5\).
   • Substitute this into the second equation: \(2x + 3(3x - 5) = 7\).
   • \(2x + 9x - 15 = 7 \implies 11x = 22 \implies x = 2\).
   • Substitute \(x=2\) back to find y: \(y = 3(2) - 5 = 1\).
   • The point of intersection is \((2, 1)\).
2. Find the Slope: The slope \(m\) of a line is the tangent of the angle it makes with the positive x-axis.
   • \(m = \tan(45^\circ) = 1\).
3. Use the Point-Slope Form: The equation of a line with slope \(m\) passing through \((x_1, y_1)\) is \(y - y_1 = m(x - x_1)\).
   • \(y - 1 = 1(x - 2)\).
4. Simplify to General Form:
   • \(y - 1 = x - 2\).
   • \(x - y - 1 = 0\).

Ответ: The equation of the line is \(x - y - 1 = 0\).

4.3. Уравнение высоты треугольника (Лаба 6, Задание 3)

\(A(4,1)\), \(B(7,4)\), and \(C(5,-2)\) are the vertices of a triangle. Find the line equation which goes from A and is perpendicular to BC.

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1. Find the Slope of BC:
   • \(m_{BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{-2 - 4}{5 - 7} = \frac{-6}{-2} = 3\).
2. Find the Slope of the Perpendicular Line: The line from A is perpendicular to BC, so its slope is the negative reciprocal of the slope of BC.
   • \(m_{\perp} = -\frac{1}{m_{BC}} = -\frac{1}{3}\).
3. Use the Point-Slope Form: The line passes through point \(A(4,1)\) with a slope of \(-1/3\).
   • \(y - 1 = -\frac{1}{3}(x - 4)\).
4. Simplify to General Form:
   • \(3(y - 1) = -(x - 4)\).
   • \(3y - 3 = -x + 4\).
   • \(x + 3y - 7 = 0\).

Ответ: The equation of the altitude from A is \(x + 3y - 7 = 0\).

4.4. Центроид треугольника (Лаба 6, Задание 4)

Find the centroid of the triangle formed by the lines given by the equations \(12x^2 - 20xy + 7y^2 = 0\) and \(2x - 3y + 4 = 0\).

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1. Find the Equations of the Three Lines:
   • The equation \(12x^2 - 20xy + 7y^2 = 0\) represents a pair of straight lines passing through the origin. We can factor it:
   • \(12x^2 - 14xy - 6xy + 7y^2 = 0 \implies 2x(6x - 7y) - y(6x - 7y) = 0 \implies (2x - y)(6x - 7y) = 0\).
   • The three lines forming the triangle are:
     • \(L_1: 2x - y = 0\)
     • \(L_2: 6x - 7y = 0\)
     • \(L_3: 2x - 3y + 4 = 0\)
2. Find the Vertices of the Triangle:
   • Vertex 1 (Intersection of \(L_1\) and \(L_2\)): Both lines pass through the origin, so their intersection is \((0, 0)\).
   • Vertex 2 (Intersection of \(L_1\) and \(L_3\)): Substitute \(y=2x\) into \(L_3\): \(2x - 3(2x) + 4 = 0 \implies -4x = -4 \implies x=1\). Then \(y=2(1)=2\). The vertex is \((1, 2)\).
   • Vertex 3 (Intersection of \(L_2\) and \(L_3\)): Substitute \(y=\frac{6}{7}x\) into \(L_3\): \(2x - 3(\frac{6}{7}x) + 4 = 0 \implies 14x - 18x + 28 = 0 \implies -4x = -28 \implies x=7\). Then \(y=\frac{6}{7}(7)=6\). The vertex is \((7, 6)\).
3. Calculate the Centroid: The centroid is the average of the coordinates of the vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\).
   • \(C_x = \frac{0 + 1 + 7}{3} = \frac{8}{3}\).
   • \(C_y = \frac{0 + 2 + 6}{3} = \frac{8}{3}\).

Ответ: The centroid of the triangle is \((\frac{8}{3}, \frac{8}{3})\).

4.5. Плоскость, параллельная данной (Лаба 6, Задание 5)

Find the equation of the plane passing through the point \((2, -3, 4)\) and parallel to the plane \(2x - 5y - 7z + 15 = 0\).

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1. Identify the Normal Vector: Parallel planes share the same normal vector. The coefficients of the variables in the plane equation give the normal vector.
   • \(\vec{n} = (2, -5, -7)\).
2. Write the General Equation of the New Plane: The equation of any plane parallel to the given one is of the form \(2x - 5y - 7z = d\).
3. Find the Constant \(d\): The plane must pass through the point \((2, -3, 4)\). Substitute these coordinates into the equation to solve for \(d\).
   • \(2(2) - 5(-3) - 7(4) = d\).
   • \(4 + 15 - 28 = d\).
   • \(d = -9\).
4. Write the Final Equation:
   • \(2x - 5y - 7z = -9\).

Ответ: The equation of the plane is \(2x - 5y - 7z + 9 = 0\).

4.6. Плоскость, перпендикулярная двум данным (Лаба 6, Задание 6)

Find the equation of the plane passing through the point \((-1, 3, 2)\) and perpendicular to the planes \(x + 2y + 2z = 5\) and \(3x + 3y + 2z = 8\).

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1. Find the Normal Vectors of the Given Planes:
   • \(\vec{n_1} = (1, 2, 2)\).
   • \(\vec{n_2} = (3, 3, 2)\).
2. Find the Normal Vector of the New Plane: The normal vector \(\vec{n}\) of the desired plane must be perpendicular to both \(\vec{n_1}\) and \(\vec{n_2}\). We can find this by calculating their cross product.
   • \(\vec{n} = \vec{n_1} \times \vec{n_2} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{pmatrix}\)
   • \(= \mathbf{i}(4 - 6) - \mathbf{j}(2 - 6) + \mathbf{k}(3 - 6) = -2\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\).
   • So, \(\vec{n} = (-2, 4, -3)\).
3. Use the Point-Normal Form: Use the calculated normal vector and the given point \((-1, 3, 2)\).
   • \(-2(x - (-1)) + 4(y - 3) - 3(z - 2) = 0\).
4. Simplify to General Form:
   • \(-2x - 2 + 4y - 12 - 3z + 6 = 0\).
   • \(-2x + 4y - 3z - 8 = 0\).
   • Multiplying by -1 gives \(2x - 4y + 3z + 8 = 0\).

Ответ: The equation of the plane is \(2x - 4y + 3z + 8 = 0\).

4.7. Отношение, в котором плоскость делит отрезок (Лаба 6, Задание 7)

Find the ratio in which the line joining the points \(P(2, -1, 4)\) and \(Q(6, 2, 4)\) is divided by the plane \(x + 2y + 3z + 5 = 0\).

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1. Use the Section Formula: Let the point of intersection R divide the line segment PQ in the ratio \(k:1\). The coordinates of R are given by \(R = (\frac{k \cdot 6 + 1 \cdot 2}{k+1}, \frac{k \cdot 2 + 1 \cdot (-1)}{k+1}, \frac{k \cdot 4 + 1 \cdot 4}{k+1})\).
2. Substitute into the Plane Equation: Since R lies on the plane, its coordinates must satisfy the plane’s equation.
   • \((\frac{6k+2}{k+1}) + 2(\frac{2k-1}{k+1}) + 3(\frac{4k+4}{k+1}) + 5 = 0\).
3. Solve for \(k\): Multiply the entire equation by \((k+1)\) to clear the denominator.
   • \((6k+2) + (4k-2) + (12k+12) + 5(k+1) = 0\).
   • \(6k + 2 + 4k - 2 + 12k + 12 + 5k + 5 = 0\).
   • \(27k + 17 = 0\).
   • \(k = -17/27\).
4. Interpret the Ratio: The ratio is \(k:1\), which is \(-17/27:1\) or \(-17:27\). The negative sign indicates that the plane divides the line segment PQ externally.

Ответ: The plane divides the line segment in the ratio -17:27 (externally).

4.8. Расстояние от точки до прямой (перпендикуляр) (Лаба 6, Задание 8)

Find the perpendicular distance from the point \(P(1, 3, -1)\) to the line \(\frac{x-13}{5} = \frac{y+8}{-8} = \frac{z-31}{1}\).

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1. Identify Information from the Line:
   • A point on the line is \(Q = (13, -8, 31)\).
   • The direction vector of the line is \(\vec{v} = (5, -8, 1)\).
2. Find the Vector \(\vec{QP}\):
   • \(\vec{QP} = P - Q = (1-13, 3-(-8), -1-31) = (-12, 11, -32)\).
3. Use the Distance Formula: The distance is \(d = \frac{||\vec{QP} \times \vec{v}||}{||\vec{v}||}\).
4. Calculate the Cross Product:
   • \(\vec{QP} \times \vec{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -12 & 11 & -32 \\ 5 & -8 & 1 \end{pmatrix}\)
   • \(= \mathbf{i}(11 - 256) - \mathbf{j}(-12 + 160) + \mathbf{k}(96 - 55) = (-245, -148, 41)\).
5. Calculate the Magnitudes:
   • \(||\vec{QP} \times \vec{v}|| = \sqrt{(-245)^2 + (-148)^2 + 41^2} = \sqrt{60025 + 21904 + 1681} = \sqrt{83610}\).
   • \(||\vec{v}|| = \sqrt{5^2 + (-8)^2 + 1^2} = \sqrt{25 + 64 + 1} = \sqrt{90}\).
6. Calculate the Distance:
   • \(d = \frac{\sqrt{83610}}{\sqrt{90}} = \sqrt{\frac{83610}{90}} = \sqrt{929}\).

Ответ: The distance is \(\sqrt{929}\).

4.9. Расстояния от точек до плоскости (Лаба 6, Задание 9)

Find the distance of the points \((2, 3, -5)\) and \((3, 4, 7)\) from the plane \(x + 2y - 2z = 9\) and prove that these points lie on the opposite sides of the plane.

Нажмите, чтобы увидеть решение

1. Rewrite the Plane Equation: Let \(F(x, y, z) = x + 2y - 2z - 9\).
2. Calculate the Distances: Use the distance formula \(d = \frac{|ax_0 + by_0 + cz_0 + D|}{\sqrt{a^2 + b^2 + c^2}}\).
   • The denominator is \(\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\).
   • For point \(P_1(2, 3, -5)\): \(d_1 = \frac{|2 + 2(3) - 2(-5) - 9|}{3} = \frac{|2 + 6 + 10 - 9|}{3} = \frac{|9|}{3} = 3\).
   • For point \(P_2(3, 4, 7)\): \(d_2 = \frac{|3 + 2(4) - 2(7) - 9|}{3} = \frac{|3 + 8 - 14 - 9|}{3} = \frac{|-12|}{3} = 4\).
3. Check if Points are on Opposite Sides: Substitute the coordinates of each point into the expression \(F(x, y, z)\). If the signs of the results are different, the points are on opposite sides.
   • \(F(2, 3, -5) = 2 + 2(3) - 2(-5) - 9 = 9\) (positive).
   • \(F(3, 4, 7) = 3 + 2(4) - 2(7) - 9 = -12\) (negative).
4. Conclusion: Since one result is positive and the other is negative, the points lie on opposite sides of the plane.

Ответ: The distance for \((2, 3, -5)\) is 3. The distance for \((3, 4, 7)\) is 4. They lie on opposite sides of the plane because substituting them into the plane’s expression yields values with opposite signs.

4.10. Угол между двумя плоскостями (Лаба 6, Задание 10)

Given two planes \(P_1: a_1x + b_1y + c_1z + d_1 = 0\) and \(P_2: a_2x + b_2y + c_2z + d_2 = 0\) where \(\vec{n_1} = (a_1, b_1, c_1)\) and \(\vec{n_2} = (a_2, b_2, c_2)\) are non-zero vectors, find the angle between these two planes.

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1. Definition: The angle between two planes is defined as the acute angle between their normal vectors.
2. Normal Vectors: The normal vectors are given by the coefficients of the variables in the plane equations:
   • \(\vec{n_1} = (a_1, b_1, c_1)\)
   • \(\vec{n_2} = (a_2, b_2, c_2)\)
3. Dot Product Formula: The cosine of the angle \(\theta\) between two vectors is given by the dot product formula: \(\cos\theta = \frac{\vec{n_1} \cdot \vec{n_2}}{||\vec{n_1}|| \cdot ||\vec{n_2}||}\).
4. Acute Angle: To ensure the angle is acute (between \(0^\circ\) and \(90^\circ\)), we take the absolute value of the dot product.
   • \(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||}\).
5. Final Formula: The angle \(\theta\) is the arccosine of this value.
   • \(\theta = \arccos\left( \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \right)\).

Ответ: The angle \(\theta\) between the two planes is given by the formula \(\theta = \arccos\left( \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||} \right)\).

4.11. Обратная матрица (вычисление) (Лаба 6, Задание 11)

Given \(A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}\), compute the inverse of A using:

1. Adjugate method
2. Gauss-Jordan method

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(a) Adjugate Method

1. Calculate the Determinant:
   • \(\det(A) = 1(4\cdot4 - 3\cdot3) - 3(1\cdot4 - 3\cdot1) + 3(1\cdot3 - 4\cdot1)\)
   • \(= 1(16-9) - 3(4-3) + 3(3-4) = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1\).
2. Find the Matrix of Cofactors:
   • \(C = \begin{bmatrix} (16-9) & -(4-3) & (3-4) \\ -(12-9) & (4-3) & -(3-3) \\ (9-12) & -(3-3) & (4-3) \end{bmatrix} = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}\).
3. Find the Adjugate Matrix: The adjugate is the transpose of the cofactor matrix.
   • \(\text{adj}(A) = C^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\).
4. Find the Inverse: \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\).
   • \(A^{-1} = \frac{1}{1}\begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\).

(b) Gauss-Jordan Method

1. Set up the Augmented Matrix \([A|I]\):
   • \(\begin{bmatrix} 1 & 3 & 3 & | & 1 & 0 & 0 \\ 1 & 4 & 3 & | & 0 & 1 & 0 \\ 1 & 3 & 4 & | & 0 & 0 & 1 \end{bmatrix}\)
2. Perform Row Operations to get RREF:
   • \(R_2 \to R_2-R_1\), \(R_3 \to R_3-R_1\): \(\begin{bmatrix} 1 & 3 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -1 & 1 & 0 \\ 0 & 0 & 1 & | & -1 & 0 & 1 \end{bmatrix}\)
   • \(R_1 \to R_1-3R_2\): \(\begin{bmatrix} 1 & 0 & 3 & | & 4 & -3 & 0 \\ 0 & 1 & 0 & | & -1 & 1 & 0 \\ 0 & 0 & 1 & | & -1 & 0 & 1 \end{bmatrix}\)
   • \(R_1 \to R_1-3R_3\): \(\begin{bmatrix} 1 & 0 & 0 & | & 7 & -3 & -3 \\ 0 & 1 & 0 & | & -1 & 1 & 0 \\ 0 & 0 & 1 & | & -1 & 0 & 1 \end{bmatrix}\)

Ответ: Both methods yield the inverse \(A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\).

4.12. Обратная матрица (второй пример) (Лаба 6, Задание 12)

Given \(B = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix}\), compute the inverse of B using:

1. Adjugate method
2. Gauss-Jordan method

Нажмите, чтобы увидеть решение

(a) Adjugate Method

1. Calculate the Determinant:
   • \(\det(B) = 1(3(-4) - (-3)(-4)) - 1(1(-4) - (-3)(-2)) + 3(1(-4) - 3(-2))\)
   • \(= 1(-12-12) - 1(-4-6) + 3(-4+6) = -24 + 10 + 6 = -8\).
2. Find the Matrix of Cofactors:
   • \(C = \begin{bmatrix} -24 & 10 & 2 \\ 8 & 2 & 2 \\ -12 & 6 & 2 \end{bmatrix}\).
3. Find the Adjugate Matrix:
   • \(\text{adj}(B) = C^T = \begin{bmatrix} -24 & 8 & -12 \\ 10 & 2 & 6 \\ 2 & 2 & 2 \end{bmatrix}\).
4. Find the Inverse: \(B^{-1} = \frac{1}{-8}\begin{bmatrix} -24 & 8 & -12 \\ 10 & 2 & 6 \\ 2 & 2 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 3/2 \\ -5/4 & -1/4 & -3/4 \\ -1/4 & -1/4 & -1/4 \end{bmatrix}\).

(b) Gauss-Jordan Method

1. Set up \([B|I]\) and reduce:
   • \(\begin{bmatrix} 1 & 1 & 3 & | & 1 & 0 & 0 \\ 1 & 3 & -3 & | & 0 & 1 & 0 \\ -2 & -4 & -4 & | & 0 & 0 & 1 \end{bmatrix} \xrightarrow{...} \begin{bmatrix} 1 & 0 & 0 & | & 3 & -1 & 3/2 \\ 0 & 1 & 0 & | & -5/4 & -1/4 & -3/4 \\ 0 & 0 & 1 & | & -1/4 & -1/4 & -1/4 \end{bmatrix}\)

Ответ: Both methods yield the inverse \(B^{-1} = \begin{bmatrix} 3 & -1 & 3/2 \\ -5/4 & -1/4 & -3/4 \\ -1/4 & -1/4 & -1/4 \end{bmatrix}\).

4.13. Ступенчатые формы: классификация (Лаба 6, Задание 13)

Classify the following matrices as either row echelon form (REF), reduced row echelon form (RREF), both, or neither.

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1. (a) \(\begin{bmatrix} 5 & 2 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 0 \end{bmatrix}\): REF. The leading entries (5, 8) are to the right of the one above. It is not RREF because the leading entries are not 1.
2. (b) \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 6 \\ 2 & 0 & 0 \end{bmatrix}\): Neither. The leading entry of row 3 (2) is not to the right of the leading entry of row 2 (1).
3. (c) \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\): REF. All zero rows are at the bottom and pivots are to the right of pivots in rows above. Not RREF because the pivot in row 1 is 2, not 1, and the pivot in column 3 has another non-zero entry (4) above it.
4. (d) \(\begin{bmatrix} 1 & 0 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\): Both REF and RREF. It satisfies all conditions for REF. It also satisfies the RREF conditions: all pivots are 1, and they are the only non-zero entries in their respective columns.
5. (e) \(\begin{bmatrix} 1 & 8 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\): Neither. The row of all zeros is not at the bottom of the matrix.
6. (f) \(\begin{bmatrix} 1 & 40 & 0 & 18 \\ 0 & 0 & 1 & -2 \end{bmatrix}\): Both REF and RREF. The pivots (the two 1s) are the only non-zero entries in their columns.

4.14. Система линейных уравнений: несколько способов (Лаба 6, Задание 14)

Given the system \(\begin{cases} x_1 + x_2 + 2x_3 = 8 \\ -x_1 - 2x_2 + 3x_3 = 1 \\ 3x_1 - 7x_2 + 4x_3 = 10 \end{cases}\), solve using:

1. Gaussian elimination
2. Gauss-Jordan elimination
3. Cramer’s rule

Нажмите, чтобы увидеть решение

(a) Gaussian Elimination

1. Row reduce the augmented matrix to REF:
   • \(\begin{bmatrix} 1 & 1 & 2 & | & 8 \\ -1 & -2 & 3 & | & 1 \\ 3 & -7 & 4 & | & 10 \end{bmatrix} \xrightarrow{...} \begin{bmatrix} 1 & 1 & 2 & | & 8 \\ 0 & -1 & 5 & | & 9 \\ 0 & 0 & -52 & | & -104 \end{bmatrix}\).
2. Back Substitution:
   • From R3: \(-52x_3 = -104 \implies x_3 = 2\).
   • From R2: \(-x_2 + 5(2) = 9 \implies -x_2 = -1 \implies x_2 = 1\).
   • From R1: \(x_1 + 1 + 2(2) = 8 \implies x_1 = 3\).

(b) Gauss-Jordan Elimination

1. Continue row reduction from REF to RREF:
   • \(\begin{bmatrix} 1 & 1 & 2 & | & 8 \\ 0 & -1 & 5 & | & 9 \\ 0 & 0 & -52 & | & -104 \end{bmatrix} \xrightarrow{...} \begin{bmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & 2 \end{bmatrix}\).
2. Read the Solution: \(x_1 = 3, x_2 = 1, x_3 = 2\).

(c) Cramer’s Rule

1. Calculate Determinants:
   • \(D = \det\begin{bmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4 \end{bmatrix} = 52\).
   • \(D_{x1} = \det\begin{bmatrix} 8 & 1 & 2 \\ 1 & -2 & 3 \\ 10 & -7 & 4 \end{bmatrix} = 156\).
   • \(D_{x2} = \det\begin{bmatrix} 1 & 8 & 2 \\ -1 & 1 & 3 \\ 3 & 10 & 4 \end{bmatrix} = 52\).
   • \(D_{x3} = \det\begin{bmatrix} 1 & 1 & 8 \\ -1 & -2 & 1 \\ 3 & -7 & 10 \end{bmatrix} = 104\).
2. Find the Solution:
   • \(x_1 = D_{x1}/D = 156/52 = 3\).
   • \(x_2 = D_{x2}/D = 52/52 = 1\).
   • \(x_3 = D_{x3}/D = 104/52 = 2\).

Ответ: The unique solution is \((3, 1, 2)\).

4.15. Однородная линейная система (Лаба 6, Задание 15)

Given the system \(\begin{cases} x_3+x_4+x_5=0 \\ -x_1-x_2-2x_3-3x_4+x_5=0 \\ x_1+x_2-2x_3-x_5=0 \\ 2x_1+x_2-2x_3+x_4+x_5=0 \end{cases}\), solve using:

1. Gaussian elimination
2. Gauss-Jordan elimination

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(a) Gaussian Elimination

1. Row reduce the coefficient matrix to REF:
   • \(\begin{bmatrix} 0 & 0 & 1 & 1 & 1 \\ -1 & -1 & -2 & -3 & 1 \\ 1 & 1 & -2 & 0 & -1 \\ 2 & 1 & -2 & 1 & 1 \end{bmatrix} \xrightarrow{...} \begin{bmatrix} 1 & 1 & -2 & 0 & -1 \\ 0 & -1 & 2 & 1 & 3 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 4 \end{bmatrix}\).
2. Back Substitution: The free variable is \(x_5\). Let \(x_5 = t\).
   • From R4: \(x_4 + 4x_5 = 0 \implies x_4 = -4t\).
   • From R3: \(x_3 + x_4 + x_5 = 0 \implies x_3 - 4t + t = 0 \implies x_3 = 3t\).
   • From R2: \(-x_2 + 2x_3 + x_4 + 3x_5 = 0 \implies -x_2 + 6t - 4t + 3t = 0 \implies x_2 = 5t\).
   • From R1: \(x_1 + x_2 - 2x_3 - x_5 = 0 \implies x_1 + 5t - 6t - t = 0 \implies x_1 = 2t\).

(b) Gauss-Jordan Elimination

1. Continue row reduction from REF to RREF:
   • \(\begin{bmatrix} 1 & 1 & -2 & 0 & -1 \\ 0 & -1 & 2 & 1 & 3 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 4 \end{bmatrix} \xrightarrow{...} \begin{bmatrix} 1 & 0 & 0 & 0 & -2 \\ 0 & 1 & 0 & 0 & -5 \\ 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 1 & 4 \end{bmatrix}\).
2. Read the Solution:
   • \(x_1 - 2x_5 = 0 \implies x_1 = 2x_5\).
   • \(x_2 - 5x_5 = 0 \implies x_2 = 5x_5\).
   • \(x_3 - 3x_5 = 0 \implies x_3 = 3x_5\).
   • \(x_4 + 4x_5 = 0 \implies x_4 = -4x_5\).
   • Let \(x_5=t\).

Ответ: The general solution is \((2t, 5t, 3t, -4t, t)\) for any scalar \(t\).

4.16. Уравнение плоскости (Лекция 6, Пример 1)

Find the equation of the plane passing through the point \((1, 2, 3)\) with a normal vector of \((2, -1, 4)\).

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1. Use the Point-Normal Form: The equation of a plane is given by \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), where \(\vec{n} = (a, b, c)\) is the normal vector and \(P_0 = (x_0, y_0, z_0)\) is a point on the plane.
2. Substitute the given values: We have \((a, b, c) = (2, -1, 4)\) and \((x_0, y_0, z_0) = (1, 2, 3)\).
   • \(2(x - 1) - 1(y - 2) + 4(z - 3) = 0\).
3. Simplify to the General Form: Expand and simplify the equation.
   • \(2x - 2 - y + 2 + 4z - 12 = 0\).
   • \(2x - y + 4z - 12 = 0\).
   • \(2x - y + 4z = 12\).

Ответ: The equation of the plane is \(2x - y + 4z = 12\).

4.17. Нормаль к плоскости (Лекция 6, Пример 2)

Given the plane \(2x - 3y + 4z = 12\), find its normal vector.

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1. Recall the General Form: The general equation of a plane is \(ax + by + cz = d\).
2. Identify the Coefficients: The coefficients of \(x, y,\) and \(z\) in the general form represent the components of a vector normal (perpendicular) to the plane.
3. Extract the Normal Vector: For the equation \(2x - 3y + 4z = 12\), the coefficients are \(a=2, b=-3, c=4\).
   • The normal vector is \(\vec{n} = (2, -3, 4)\).

Ответ: A normal vector to the plane is \(\vec{n} = (2, -3, 4)\).

4.18. Прямая пересечения двух плоскостей (Лекция 6, Пример 3)

Find the line of intersection of the planes: \[ \begin{cases} x + 2y - z = 4 \\ 2x - y + 3z = 1 \end{cases} \]

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1. Find the Direction Vector: The line of intersection lies in both planes, so its direction vector \(\vec{v}\) must be perpendicular to both normal vectors, \(\vec{n_1} = (1, 2, -1)\) and \(\vec{n_2} = (2, -1, 3)\). We can find \(\vec{v}\) by taking the cross product.
   • \(\vec{v} = \vec{n_1} \times \vec{n_2} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{pmatrix}\)
   • \(= \mathbf{i}(2\cdot3 - (-1)\cdot(-1)) - \mathbf{j}(1\cdot3 - (-1)\cdot2) + \mathbf{k}(1\cdot(-1) - 2\cdot2)\)
   • \(= \mathbf{i}(6-1) - \mathbf{j}(3+2) + \mathbf{k}(-1-4) = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k} = (5, -5, -5)\). We can use a simpler parallel vector, like \((1, -1, -1)\).
2. Find a Point on the Line: To find a point, we can set one variable to zero and solve the resulting system for the other two. Let’s set \(z=0\).
   • \(x + 2y = 4\)
   • \(2x - y = 1\)
3. Solve the 2x2 System:
   • From the second equation, \(y = 2x - 1\).
   • Substitute into the first: \(x + 2(2x - 1) = 4 \implies x + 4x - 2 = 4 \implies 5x = 6 \implies x = 6/5\).
   • Now find \(y\): \(y = 2(6/5) - 1 = 12/5 - 5/5 = 7/5\).
   • So, a point on the line is \(P_0 = (6/5, 7/5, 0)\).
4. Write the Equation of the Line: Using the point and the direction vector \(\vec{v} = (1, -1, -1)\), the symmetric equation is:
   • \(\frac{x - 6/5}{1} = \frac{y - 7/5}{-1} = \frac{z}{-1}\).

Ответ: The line of intersection is given by the symmetric equations \(\frac{x - 6/5}{1} = \frac{y - 7/5}{-1} = \frac{z}{-1}\).

4.19. Операции с векторами (Лекция 6, Пример 4)

Given vectors \(\vec{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(\vec{b} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), find:

1. The dot product \(\vec{a} \cdot \vec{b}\)
2. The cross product \(\vec{a} \times \vec{b}\)
3. The angle between \(\vec{a}\) and \(\vec{b}\)

Нажмите, чтобы увидеть решение

First, write the vectors in component form: \(\vec{a} = (2, -1, 3)\) and \(\vec{b} = (-1, 2, 1)\).

(a) The dot product \(\vec{a} \cdot \vec{b}\)

1. Use the dot product formula: \(\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\).
2. Calculate:
   • \(\vec{a} \cdot \vec{b} = (2)(-1) + (-1)(2) + (3)(1) = -2 - 2 + 3 = -1\).
   • Answer (a): \(-1\).

(b) The cross product \(\vec{a} \times \vec{b}\)

1. Set up the determinant:
   • \(\vec{a} \times \vec{b} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ -1 & 2 & 1 \end{pmatrix}\)
2. Calculate:
   • \(= \mathbf{i}((-1)(1) - (3)(2)) - \mathbf{j}((2)(1) - (3)(-1)) + \mathbf{k}((2)(2) - (-1)(-1))\)
   • \(= \mathbf{i}(-1 - 6) - \mathbf{j}(2 + 3) + \mathbf{k}(4 - 1) = -7\mathbf{i} - 5\mathbf{j} + 3\mathbf{k}\).
   • Answer (b): \((-7, -5, 3)\).

(c) The angle between \(\vec{a}\) and \(\vec{b}\)

1. Calculate the magnitudes:
   • \(||\vec{a}|| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}\).
   • \(||\vec{b}|| = \sqrt{(-1)^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\).
2. Use the angle formula: \(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| \cdot ||\vec{b}||}\).
3. Calculate:
   • \(\cos\theta = \frac{-1}{\sqrt{14}\sqrt{6}} = \frac{-1}{\sqrt{84}}\).
   • \(\theta = \arccos\left(\frac{-1}{\sqrt{84}}\right) \approx 96.28^\circ\).
   • Answer (c): \(\arccos\left(\frac{-1}{\sqrt{84}}\right)\).

4.20. Плоскость через три точки (Лекция 6, Пример 5)

Find the equation of the plane passing through the points \((1, 2, 3)\), \((2, -1, 1)\), and \((3, 1, -2)\).

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1. Find Two Vectors in the Plane: Let the points be \(P_1, P_2, P_3\).
   • \(\vec{P_1P_2} = (2-1, -1-2, 1-3) = (1, -3, -2)\).
   • \(\vec{P_1P_3} = (3-1, 1-2, -2-3) = (2, -1, -5)\).
2. Find the Normal Vector: The normal vector \(\vec{n}\) is the cross product of these two vectors.
   • \(\vec{n} = \vec{P_1P_2} \times \vec{P_1P_3} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & -2 \\ 2 & -1 & -5 \end{pmatrix}\)
   • \(= \mathbf{i}(15 - 2) - \mathbf{j}(-5 - (-4)) + \mathbf{k}(-1 - (-6)) = 13\mathbf{i} + \mathbf{j} + 5\mathbf{k}\).
   • So, \(\vec{n} = (13, 1, 5)\).
3. Use the Point-Normal Form: Using point \(P_1(1, 2, 3)\) and \(\vec{n}=(13, 1, 5)\).
   • \(13(x-1) + 1(y-2) + 5(z-3) = 0\).
4. Simplify:
   • \(13x - 13 + y - 2 + 5z - 15 = 0\).
   • \(13x + y + 5z = 30\).

Ответ: The equation of the plane is \(13x + y + 5z = 30\).

4.21. Параллельны ли прямые? (Лекция 6, Пример 6)

Consider the two lines: \(L_1: \frac{x - 1}{-2} = \frac{y + 1}{1} = \frac{z - 1}{2}\) \(L_2: \frac{x}{1} = \frac{x - 1}{-1/2} = \frac{z}{-1}\) Explain why they are parallel.

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1. Identify the Direction Vectors: The direction vector of a line in symmetric form is given by the denominators of the fractions.
   • Direction vector of \(L_1\) is \(\vec{v_1} = (-2, 1, 2)\).
   • Direction vector of \(L_2\) is \(\vec{v_2} = (1, -1/2, -1)\).
2. Check for Proportionality: Two lines are parallel if their direction vectors are scalar multiples of each other, i.e., \(\vec{v_1} = k\vec{v_2}\) for some scalar \(k\).
3. Compare the Vectors: Let’s see if we can find a \(k\) such that \((-2, 1, 2) = k(1, -1/2, -1)\).
   • From the first component: \(-2 = k(1) \implies k = -2\).
   • Check with the other components:
     • Is \(1 = (-2)(-1/2)\)? Yes, \(1 = 1\).
     • Is \(2 = (-2)(-1)\)? Yes, \(2 = 2\).
4. Conclusion: Since \(\vec{v_1} = -2\vec{v_2}\), the direction vectors are parallel, and therefore the lines are parallel.

Ответ: The lines are parallel because their direction vectors, \((-2, 1, 2)\) and \((1, -1/2, -1)\), are scalar multiples of each other.

4.22. Смешанное произведение (Лекция 6, Пример 7)

Find the scalar triple product of vectors \(\vec{a} = (1, 2, 3)\), \(\vec{b} = (2, -1, 1)\), and \(\vec{c} = (3, 1, -2)\). What does the result tell you about these vectors?

Нажмите, чтобы увидеть решение

1. Set up the determinant: The scalar triple product \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is the determinant of the matrix formed by the three vectors.
   • \(\det = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{vmatrix}\)
2. Calculate the determinant:
   • \(= 1((-1)(-2) - 1(1)) - 2((2)(-2) - 1(3)) + 3((2)(1) - (-1)(3))\)
   • \(= 1(2 - 1) - 2(-4 - 3) + 3(2 + 3)\)
   • \(= 1(1) - 2(-7) + 3(5) = 1 + 14 + 15 = 30\).
3. Interpret the result: The scalar triple product represents the volume of the parallelepiped formed by the three vectors. Since the result is non-zero, the vectors are not coplanar (they do not lie in the same plane).

Ответ: The scalar triple product is 30. This means the vectors are not coplanar.

4.23. Прямые: параллельны, пересекаются или скрещиваются? (Лекция 6, Пример 8)

Determine whether the lines \(\frac{x-1}{2} = \frac{x-2}{3} = \frac{z-3}{1}\) and \(\frac{x-4}{1} = \frac{y-1}{3} = \frac{z-5}{-2}\) are parallel, intersecting, or skew.

Нажмите, чтобы увидеть решение

1. Check for Parallelism: Compare the direction vectors.
   • \(\vec{v_1} = (2, 3, 1)\).
   • \(\vec{v_2} = (1, 3, -2)\).
   • The vectors are not scalar multiples of each other, so the lines are not parallel.
2. Check for Intersection: Write the lines in parametric form and set the coordinates equal.
   • \(L_1\): \(x=1+2t, y=2+3t, z=3+t\).
   • \(L_2\): \(x=4+s, y=1+3s, z=5-2s\).
3. Set up and solve the system:
   • \(1+2t = 4+s \implies s = 2t-3\).
   • \(2+3t = 1+3s\).
   • \(3+t = 5-2s\).
   • Substitute \(s=2t-3\) into the second equation: \(2+3t = 1+3(2t-3) \implies 2+3t = 1+6t-9 \implies 2+3t = 6t-8 \implies 10 = 3t \implies t=10/3\).
   • Find \(s\): \(s = 2(10/3)-3 = 20/3 - 9/3 = 11/3\).
   • Check these values in the third equation: \(3+(10/3) = 5-2(11/3) \implies 19/3 = 15/3 - 22/3 \implies 19/3 = -7/3\). This is a contradiction.
4. Conclusion: The lines are not parallel and they do not intersect.

Ответ: The lines are skew.

4.24. Параллельны ли прямые? (второй пример) (Лекция 6, Пример 9)

Consider the following lines: \(L_1: \frac{x - 2}{2} = \frac{y + 2}{1} = \frac{z - 2}{-1}\) \(L_2: \frac{x - 1}{1} = \frac{y}{-2} = \frac{z}{2}\) Explain why they are not parallel.

Нажмите, чтобы увидеть решение

1. Identify the Direction Vectors:
   • Direction vector of \(L_1\) is \(\vec{v_1} = (2, 1, -1)\).
   • Direction vector of \(L_2\) is \(\vec{v_2} = (1, -2, 2)\).
2. Check for Proportionality: Two lines are parallel if their direction vectors are scalar multiples of each other. We check if there is a scalar \(k\) such that \(\vec{v_1} = k\vec{v_2}\).
3. Compare the Components:
   • From the x-component: \(2 = k(1) \implies k = 2\).
   • Check this \(k\) with the y-component: Is \(1 = (2)(-2)\)? No, \(1 \neq -4\).
4. Conclusion: Since the components are not proportional, the direction vectors are not scalar multiples of each other.

Ответ: The lines are not parallel because their direction vectors, \((2, 1, -1)\) and \((1, -2, 2)\), are not scalar multiples of each other.

4.25. Проекция вектора (Лекция 6, Пример 10)

Find the projection of vector \(\vec{v} = (4, 2, 1)\) onto vector \(\vec{u} = (1, 1, 1)\).

Нажмите, чтобы увидеть решение

1. Use the projection formula: \(proj_{\vec{u}}(\vec{v}) = \frac{\vec{v} \cdot \vec{u}}{||\vec{u}||^2} \vec{u}\).
2. Calculate the dot product:
   • \(\vec{v} \cdot \vec{u} = (4)(1) + (2)(1) + (1)(1) = 7\).
3. Calculate the squared magnitude of \(\vec{u}\):
   • \(||\vec{u}||^2 = 1^2 + 1^2 + 1^2 = 3\).
4. Calculate the projection vector:
   • \(proj_{\vec{u}}(\vec{v}) = \frac{7}{3}(1, 1, 1) = (\frac{7}{3}, \frac{7}{3}, \frac{7}{3})\).

Ответ: The projection of \(\vec{v}\) onto \(\vec{u}\) is \((\frac{7}{3}, \frac{7}{3}, \frac{7}{3})\).

4.26. Расстояние от точки до плоскости (Лекция 6, Пример 11)

Find the distance from the point \((2, -1, 3)\) to the plane \(2x - 3y + 6z = 12\).

Нажмите, чтобы увидеть решение

1. State the Distance Formula: The distance from \((x_0, y_0, z_0)\) to \(ax+by+cz-d=0\) is \(D = \frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}\).
2. Identify the values:
   • Point: \((x_0, y_0, z_0) = (2, -1, 3)\).
   • Plane: \(2x - 3y + 6z - 12 = 0\), so \(a=2, b=-3, c=6, d=12\).
3. Substitute and calculate:
   • \(D = \frac{|2(2) - 3(-1) + 6(3) - 12|}{\sqrt{2^2+(-3)^2+6^2}} = \frac{|4+3+18-12|}{\sqrt{4+9+36}} = \frac{|13|}{\sqrt{49}} = \frac{13}{7}\).

Ответ: The distance is \(\frac{13}{7}\).

4.27. Скрещиваются ли прямые? (Лекция 6, Пример 12)

Consider two lines: \(L_1: \frac{x + 1}{2} = \frac{y}{-1} = \frac{z + 2}{-1}\) \(L_2: \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\) Explain why they are skew.

Нажмите, чтобы увидеть решение

1. Check for Parallelism:
   • Direction vector of \(L_1\): \(\vec{v_1} = (2, -1, -1)\).
   • Direction vector of \(L_2\): \(\vec{v_2} = (1, 2, 3)\).
   • The vectors are not scalar multiples of each other, so the lines are not parallel.
2. Check for Intersection: Two lines are skew if they are not parallel and do not intersect. We write the lines in parametric form and see if we can find a common point.
   • \(L_1\): \(x = -1 + 2t\), \(y = -t\), \(z = -2 - t\).
   • \(L_2\): \(x = s\), \(y = 2s\), \(z = 3s\).
3. Set the coordinates equal to find a potential intersection:
   • \(-1 + 2t = s\)
   • \(-t = 2s\)
   • \(-2 - t = 3s\)
4. Solve the system of equations:
   • From the second equation, \(t = -2s\).
   • Substitute this into the first equation: \(-1 + 2(-2s) = s \implies -1 - 4s = s \implies -1 = 5s \implies s = -1/5\).
   • Using \(s = -1/5\), we find \(t = -2(-1/5) = 2/5\).
   • Now check if these values of \(s\) and \(t\) satisfy the third equation: Is \(-2 - (2/5) = 3(-1/5)\)?
   • \(-12/5 = -3/5\). This is a contradiction.
5. Conclusion: Since the system of equations has no solution, the lines do not intersect. Because they are not parallel and do not intersect, they are skew.

Ответ: The lines are skew because their direction vectors are not parallel and there is no point of intersection.

4.28. Отражение вектора (Лекция 6, Пример 13)

Find the reflection of vector \(\vec{v} = (3, -2, 1)\) about the vector \(\vec{n} = (1, 2, -2)\).

Нажмите, чтобы увидеть решение

1. State the Formula: The reflection of a vector \(\vec{v}\) across the line defined by vector \(\vec{n}\) is given by the formula \(ref_{\vec{n}}(\vec{v}) = 2 \cdot proj_{\vec{n}}(\vec{v}) - \vec{v}\).
2. Calculate the projection of \(\vec{v}\) onto \(\vec{n}\):
   • \(\vec{v} \cdot \vec{n} = (3)(1) + (-2)(2) + (1)(-2) = 3 - 4 - 2 = -3\).
   • \(||\vec{n}||^2 = 1^2 + 2^2 + (-2)^2 = 1 + 4 + 4 = 9\).
   • \(proj_{\vec{n}}(\vec{v}) = \frac{-3}{9}(1, 2, -2) = -\frac{1}{3}(1, 2, -2) = (-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})\).
3. Calculate the reflection vector:
   • \(ref_{\vec{n}}(\vec{v}) = 2(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) - (3, -2, 1)\)
   • \(= (-\frac{2}{3}, -\frac{4}{3}, \frac{4}{3}) - (3, -2, 1)\)
   • \(= (-\frac{2}{3} - \frac{9}{3}, -\frac{4}{3} + \frac{6}{3}, \frac{4}{3} - \frac{3}{3}) = (-\frac{11}{3}, \frac{2}{3}, \frac{1}{3})\).

Ответ: The reflection is \((-\frac{11}{3}, \frac{2}{3}, \frac{1}{3})\).

4.29. Острый угол между плоскостями (Лекция 6, Пример 14)

Find the acute angle between the planes \(x + 2y - z = 5\) and \(3x - y + 2z = 7\).

Нажмите, чтобы увидеть решение

1. Find the Normal Vectors:
   • \(\vec{n_1} = (1, 2, -1)\).
   • \(\vec{n_2} = (3, -1, 2)\).
2. Use the Dot Product Formula: \(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||}\).
3. Calculate the components:
   • \(\vec{n_1} \cdot \vec{n_2} = 1(3) + 2(-1) + (-1)(2) = 3 - 2 - 2 = -1\).
   • \(||\vec{n_1}|| = \sqrt{1^2+2^2+(-1)^2} = \sqrt{6}\).
   • \(||\vec{n_2}|| = \sqrt{3^2+(-1)^2+2^2} = \sqrt{14}\).
4. Calculate the angle:
   • \(\cos\theta = \frac{|-1|}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}}\).
   • \(\theta = \arccos(\frac{1}{\sqrt{84}}) \approx 83.7^\circ\).

Ответ: The acute angle is \(\arccos(\frac{1}{\sqrt{84}})\).

4.30. Угол между двумя плоскостями (Лекция 6, Пример 15)

Find the angle between the planes \(x + 2y - z = 4\) and \(2x - y + 3z = 1\).

Нажмите, чтобы увидеть решение

1. Identify the Normal Vectors: The angle between two planes is the angle between their normal vectors.
   • Normal vector of the first plane: \(\vec{n_1} = (1, 2, -1)\).
   • Normal vector of the second plane: \(\vec{n_2} = (2, -1, 3)\).
2. Use the Dot Product Formula: The cosine of the angle \(\theta\) between two vectors is given by \(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||}\).
3. Calculate the Dot Product:
   • \(\vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(-1) + (-1)(3) = 2 - 2 - 3 = -3\).
4. Calculate the Magnitudes:
   • \(||\vec{n_1}|| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\).
   • \(||\vec{n_2}|| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}\).
5. Calculate the Cosine of the Angle:
   • \(\cos\theta = \frac{|-3|}{\sqrt{6}\sqrt{14}} = \frac{3}{\sqrt{84}}\).
   • \(\theta = \arccos\left(\frac{3}{\sqrt{84}}\right) \approx 70.89°\).

Ответ: The angle between the planes is \(\arccos\left(\frac{3}{\sqrt{84}}\right)\), which is approximately \(70.89°\).

4.31. Площадь треугольника и объём параллелепипеда (Лекция 6, Пример 16)

Given points \(A(1, 2, 3)\), \(B(2, -1, 1)\), and \(C(3, 1, -2)\), find:

1. The area of triangle \(ABC\)
2. The volume of the parallelepiped formed by vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\) where \(D(0, 1, 2)\)

Нажмите, чтобы увидеть решение

First, find the vectors originating from point A:

• \(\vec{AB} = B - A = (1, -3, -2)\)
• \(\vec{AC} = C - A = (2, -1, -5)\)
• \(\vec{AD} = D - A = (-1, -1, -1)\)

(a) Area of triangle \(ABC\)

1. Calculate the cross product: Area \(= \frac{1}{2}||\vec{AB} \times \vec{AC}||\).
   • \(\vec{AB} \times \vec{AC} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & -2 \\ 2 & -1 & -5 \end{pmatrix} = (13, 1, 5)\).
2. Find the magnitude:
   • \(||\vec{AB} \times \vec{AC}|| = \sqrt{13^2 + 1^2 + 5^2} = \sqrt{169 + 1 + 25} = \sqrt{195}\).
3. Calculate the area:
   • Area \(= \frac{1}{2}\sqrt{195}\).
   • Answer (a): \(\frac{\sqrt{195}}{2}\).

(b) Volume of the parallelepiped

1. Calculate the scalar triple product: Volume \(= |\vec{AD} \cdot (\vec{AB} \times \vec{AC})|\).
2. Calculate:
   • \(= |(-1, -1, -1) \cdot (13, 1, 5)|\)
   • \(= |(-1)(13) + (-1)(1) + (-1)(5)| = |-13 - 1 - 5| = |-19| = 19\).
   • Answer (b): \(19\).

4.32. Канонические уравнения прямой (Лекция 6, Пример 17)

Find the symmetric equations of the line through \((1, -2, 3)\) perpendicular to the plane \(2x + y - z = 4\).

Нажмите, чтобы увидеть решение

1. Find the Direction Vector: A line perpendicular to a plane has a direction vector parallel to the plane’s normal vector.
   • Normal vector of the plane: \(\vec{n} = (2, 1, -1)\).
   • Direction vector of the line: \(\vec{v} = (2, 1, -1)\).
2. Use the Symmetric Form: The symmetric equations for a line through \((x_0, y_0, z_0)\) with direction \((a, b, c)\) are \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).
3. Substitute the values:
   • Point: \((1, -2, 3)\).
   • Direction: \((2, 1, -1)\).
   • \(\frac{x-1}{2} = \frac{y-(-2)}{1} = \frac{z-3}{-1}\).

Ответ: The symmetric equations are \(\frac{x-1}{2} = \frac{y+2}{1} = \frac{z-3}{-1}\).

4.33. Линейная независимость векторов (Лекция 6, Пример 18)

Determine if the vectors \(\vec{a} = (1, 2, 3)\), \(\vec{b} = (2, 5, 7)\), and \(\vec{c} = (1, 3, 5)\) are linearly independent.

Нажмите, чтобы увидеть решение

1. Use the determinant: Three vectors in \(\mathbb{R}^3\) are linearly independent if and only if the determinant of the matrix formed by them is non-zero.
2. Set up the determinant:
   • \(\det = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 7 & 5 \end{vmatrix}\)
3. Calculate the determinant:
   • \(= 1(25 - 21) - 2(10 - 9) + 1(14 - 15) = 1(4) - 2(1) + 1(-1) = 4 - 2 - 1 = 1\).
4. Conclusion: Since the determinant is 1 (non-zero), the vectors are linearly independent.

Ответ: Yes, the vectors are linearly independent.

4.34. Пересечение прямой и плоскости (Лекция 6, Пример 19)

Determine the point of intersection of the line \(\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-2}{3}\) with the plane \(x + 2y - z = 6\).

Нажмите, чтобы увидеть решение

1. Write the Line in Parametric Form:
   • Let \(\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-2}{3} = t\).
   • \(x = 1 + 2t\)
   • \(y = -1 + t\)
   • \(z = 2 + 3t\)
2. Substitute into the Plane Equation:
   • \((1+2t) + 2(-1+t) - (2+3t) = 6\).
3. Solve for \(t\):
   • \(1 + 2t - 2 + 2t - 2 - 3t = 6\).
   • \(t - 3 = 6 \implies t = 9\).
4. Find the Intersection Point: Substitute \(t=9\) back into the parametric equations.
   • \(x = 1 + 2(9) = 19\).
   • \(y = -1 + 9 = 8\).
   • \(z = 2 + 3(9) = 29\).

Ответ: The point of intersection is \((19, 8, 29)\).

4.35. Расстояние от точки до плоскости (Лекция 6, Пример 20)

Find the distance from the point \((1, 2, 3)\) to the plane \(2x - y + 2z - 4 = 0\).

Нажмите, чтобы увидеть решение

1. State the Distance Formula: The distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane \(ax + by + cz + D = 0\) is given by:
   • \(d = \frac{|ax_0 + by_0 + cz_0 + D|}{\sqrt{a^2 + b^2 + c^2}}\).
2. Identify the Values:
   • Point: \((x_0, y_0, z_0) = (1, 2, 3)\).
   • Plane coefficients: \(a=2, b=-1, c=2, D=-4\).
3. Substitute into the Formula:
   • \(d = \frac{|2(1) - 1(2) + 2(3) - 4|}{\sqrt{2^2 + (-1)^2 + 2^2}}\).
4. Calculate the Result:
   • \(d = \frac{|2 - 2 + 6 - 4|}{\sqrt{4 + 1 + 4}} = \frac{|2|}{\sqrt{9}} = \frac{2}{3}\).

Ответ: The distance is \(2/3\).

4.36. Единичный вектор, перпендикулярный двум данным (Лекция 6, Пример 21)

Find a unit vector perpendicular to both \(\vec{u} = (1, 1, 0)\) and \(\vec{v} = (2, -1, 1)\).

Нажмите, чтобы увидеть решение

1. Find the cross product: A vector perpendicular to both \(\vec{u}\) and \(\vec{v}\) is their cross product.
   • \(\vec{w} = \vec{u} \times \vec{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 2 & -1 & 1 \end{pmatrix} = (1, -1, -3)\).
2. Find the magnitude of the cross product:
   • \(||\vec{w}|| = \sqrt{1^2 + (-1)^2 + (-3)^2} = \sqrt{1 + 1 + 9} = \sqrt{11}\).
3. Normalize the vector: Divide the vector by its magnitude to get a unit vector.
   • \(\hat{w} = \frac{\vec{w}}{||\vec{w}||} = \frac{1}{\sqrt{11}}(1, -1, -3) = (\frac{1}{\sqrt{11}}, -\frac{1}{\sqrt{11}}, -\frac{3}{\sqrt{11}})\).

Ответ: The unit vector is \((\frac{1}{\sqrt{11}}, -\frac{1}{\sqrt{11}}, -\frac{3}{\sqrt{11}})\).

4.37. Плоскость, содержащая две параллельные прямые (Лекция 6, Пример 22)

Find the equation of the plane containing the parallel lines: \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{-1}\) and \(\frac{x}{2} = \frac{y-1}{1} = \frac{z-2}{-1}\).

Нажмите, чтобы увидеть решение

1. Find a Direction Vector: The common direction vector is \(\vec{v} = (2, 1, -1)\).
2. Find a Vector Connecting the Lines: Pick one point from each line.
   • \(P_1 = (1, 2, 3)\) from the first line.
   • \(P_2 = (0, 1, 2)\) from the second line.
   • The vector connecting them is \(\vec{u} = \vec{P_1P_2} = (0-1, 1-2, 2-3) = (-1, -1, -1)\).
3. Find the Normal Vector to the Plane: The normal vector is the cross product of the two vectors in the plane.
   • \(\vec{n} = \vec{u} \times \vec{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & -1 \\ 2 & 1 & -1 \end{pmatrix} = (2, -3, 1)\).
4. Write the Plane Equation: Use the point-normal form with point \(P_1\) and normal \(\vec{n}\).
   • \(2(x-1) - 3(y-2) + 1(z-3) = 0\).
   • \(2x - 2 - 3y + 6 + z - 3 = 0\).
   • \(2x - 3y + z + 1 = 0\).

Ответ: The equation of the plane is \(2x - 3y + z = -1\).

4.38. Расстояние от точки до прямой (Лекция 6, Пример 23)

Compute the distance of the point \(P(5, -6, 2)\) from a line with parametric equations: \[ \begin{cases} x = 1 \\ y = -1 + 4t \\ z = 2 - 3t \end{cases}, t \in \mathbb{R} \]

Нажмите, чтобы увидеть решение

1. State the Distance Formula: The distance from a point \(P\) to a line is given by \(d = \frac{||\vec{QP} \times \vec{v}||}{||\vec{v}||}\), where \(Q\) is any point on the line and \(\vec{v}\) is the direction vector of the line.
2. Identify a Point and Direction Vector from the Line:
   • By setting \(t=0\), we get a point on the line: \(Q = (1, -1, 2)\).
   • The direction vector is given by the coefficients of \(t\): \(\vec{v} = (0, 4, -3)\).
3. Find the Vector \(\vec{QP}\):
   • \(\vec{QP} = P - Q = (5-1, -6-(-1), 2-2) = (4, -5, 0)\).
4. Calculate the Cross Product \(\vec{QP} \times \vec{v}\):
   • \(\vec{QP} \times \vec{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -5 & 0 \\ 0 & 4 & -3 \end{pmatrix}\)
   • \(= \mathbf{i}((-5)(-3) - 0(4)) - \mathbf{j}(4(-3) - 0(0)) + \mathbf{k}(4(4) - (-5)(0))\)
   • \(= 15\mathbf{i} + 12\mathbf{j} + 16\mathbf{k} = (15, 12, 16)\).
5. Calculate the Magnitudes:
   • \(||\vec{QP} \times \vec{v}|| = \sqrt{15^2 + 12^2 + 16^2} = \sqrt{225 + 144 + 256} = \sqrt{625} = 25\).
   • \(||\vec{v}|| = \sqrt{0^2 + 4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\).
6. Calculate the Distance:
   • \(d = \frac{25}{5} = 5\).

Ответ: The distance from the point to the line is \(5\).

4.39. Проверка векторного тождества (Лекция 6, Пример 24)

Verify the vector identity \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\) for vectors \(\vec{a} = (1, 2, 3)\), \(\vec{b} = (2, -1, 1)\), \(\vec{c} = (3, 1, -2)\).

Нажмите, чтобы увидеть решение

1. Calculate the Left-Hand Side (LHS):
   • First, find \(\vec{b} \times \vec{c} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{pmatrix} = (1, 7, 5)\).
   • Then, calculate \(\vec{a} \times (\vec{b} \times \vec{c}) = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & 7 & 5 \end{pmatrix} = (10-21, 3-5, 7-2) = (-11, -2, 5)\).
2. Calculate the Right-Hand Side (RHS):
   • First, calculate the dot products:
     • \(\vec{a} \cdot \vec{c} = (1)(3) + (2)(1) + (3)(-2) = 3 + 2 - 6 = -1\).
     • \(\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3\).
   • Then, calculate the full expression: \((\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = (-1)\vec{b} - (3)\vec{c}\).
   • \(= -1(2, -1, 1) - 3(3, 1, -2) = (-2, 1, -1) - (9, 3, -6) = (-11, -2, 5)\).
3. Compare: The LHS \((-11, -2, 5)\) is equal to the RHS \((-11, -2, 5)\).

Ответ: The identity is verified.

4.40. Расстояние между скрещивающимися прямыми (Лекция 6, Пример 25)

Find the distance between the skew lines: \(L_1: x = 1+t, y=t, z=1+2t\) and \(L_2: x=2+s, y=-1, z=3+s\).

Нажмите, чтобы увидеть решение

1. Identify Points and Direction Vectors:
   • \(L_1\): \(P_1=(1,0,1)\), \(\vec{v_1}=(1,1,2)\).
   • \(L_2\): \(P_2=(2,-1,3)\), \(\vec{v_2}=(1,0,1)\).
2. Find the Common Normal Vector: \(\vec{n} = \vec{v_1} \times \vec{v_2}\).
   • \(\vec{n} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{pmatrix} = (1, 1, -1)\).
3. Find the Vector Connecting the Lines: \(\vec{P_1P_2} = (2-1, -1-0, 3-1) = (1, -1, 2)\).
4. Use the Distance Formula: \(d = \frac{|\vec{P_1P_2} \cdot \vec{n}|}{||\vec{n}||}\).
5. Calculate:
   • \(\vec{P_1P_2} \cdot \vec{n} = (1)(1) + (-1)(1) + (2)(-1) = 1 - 1 - 2 = -2\).
   • \(||\vec{n}|| = \sqrt{1^2+1^2+(-1)^2} = \sqrt{3}\).
   • \(d = \frac{|-2|}{\sqrt{3}} = \frac{2}{\sqrt{3}}\).

Ответ: The distance is \(\frac{2\sqrt{3}}{3}\).

4.41. Расстояние от точки до прямой (Лекция 6, Пример 26)

Find the distance from the point \(P(2, -1, 3)\) to the line through points \(A(1, 0, 1)\) and \(B(3, 2, 0)\).

Нажмите, чтобы увидеть решение

1. Find the direction vector of the line: \(\vec{AB} = B - A = (2, 2, -1)\).
2. Find the vector from a point on the line to P: \(\vec{AP} = P - A = (1, -1, 2)\).
3. Use the distance formula: \(d = \frac{||\vec{AP} \times \vec{AB}||}{||\vec{AB}||}\).
4. Calculate the cross product:
   • \(\vec{AP} \times \vec{AB} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 2 & 2 & -1 \end{pmatrix} = (-3, 5, 4)\).
5. Calculate the magnitudes:
   • \(||\vec{AP} \times \vec{AB}|| = \sqrt{(-3)^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}\).
   • \(||\vec{AB}|| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\).
6. Calculate the distance: \(d = \frac{5\sqrt{2}}{3}\).

Ответ: The distance is \(\frac{5\sqrt{2}}{3}\).

4.42. Симметрия точки относительно плоскости (Лекция 6, Пример 27)

Find the reflection of the point \((2, 1, -3)\) in the plane \(x - 2y + 2z = 6\).

Нажмите, чтобы увидеть решение

1. Find the Line Through the Point Perpendicular to the Plane:
   • The direction vector is the plane’s normal: \(\vec{v} = (1, -2, 2)\).
   • The line is \(L: x=2+t, y=1-2t, z=-3+2t\).
2. Find the Intersection of the Line and Plane: Substitute the line’s equations into the plane’s equation.
   • \((2+t) - 2(1-2t) + 2(-3+2t) = 6\).
   • \(2+t - 2+4t - 6+4t = 6 \implies 9t - 6 = 6 \implies 9t = 12 \implies t = 4/3\).
   • The intersection point \(Q\) is: \(x=2+4/3=10/3\), \(y=1-8/3=-5/3\), \(z=-3+8/3=-1/3\). So \(Q=(\frac{10}{3}, -\frac{5}{3}, -\frac{1}{3})\).
3. Find the Reflection: The intersection \(Q\) is the midpoint of the original point \(P\) and its reflection \(P'\).
   • Let \(P'=(x', y', z')\). Then \(\frac{2+x'}{2} = \frac{10}{3}\), \(\frac{1+y'}{2} = -\frac{5}{3}\), \(\frac{-3+z'}{2} = -\frac{1}{3}\).
   • \(6+3x' = 20 \implies 3x' = 14 \implies x' = 14/3\).
   • \(3+3y' = -10 \implies 3y' = -13 \implies y' = -13/3\).
   • \(-9+3z' = -2 \implies 3z' = 7 \implies z' = 7/3\).

Ответ: The reflection is \((\frac{14}{3}, -\frac{13}{3}, \frac{7}{3})\).

4.43. Компланарны ли четыре точки? (Лекция 6, Пример 28)

Determine whether the four points \((1,1,1), (2,3,4), (3,4,5), (1,0,1)\) are coplanar.

Нажмите, чтобы увидеть решение

1. Form Three Vectors: Create three vectors from the points, using \((1,0,1)\) as the origin.
   • \(\vec{v_1} = (1-1, 1-0, 1-1) = (0, 1, 0)\).
   • \(\vec{v_2} = (2-1, 3-0, 4-1) = (1, 3, 3)\).
   • \(\vec{v_3} = (3-1, 4-0, 5-1) = (2, 4, 4)\).
2. Calculate the Scalar Triple Product: The points are coplanar if the volume of the parallelepiped formed by these vectors is zero.
   • Volume \(= \left| \det \begin{pmatrix} 0 & 1 & 0 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix} \right|\)
   • \(= |0 - 1(1\cdot4 - 3\cdot2) + 0| = |-(4-6)| = |2| = 2\).
3. Conclusion: Since the volume is not zero, the vectors are not coplanar.

Ответ: The points are not coplanar.

4.44. Определитель матрицы (Лекция 6, Пример 29)

Find the determinant of the matrix: \(A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 3 & 5 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Use the cofactor expansion method along the first row:
   • \(\det(A) = 1\begin{vmatrix} 5 & 7 \\ 3 & 5 \end{vmatrix} - 2\begin{vmatrix} 2 & 7 \\ 1 & 5 \end{vmatrix} + 3\begin{vmatrix} 2 & 5 \\ 1 & 3 \end{vmatrix}\)
2. Calculate the determinants of the \(2 \times 2\) matrices:
   • \(= 1(5 \cdot 5 - 7 \cdot 3) - 2(2 \cdot 5 - 7 \cdot 1) + 3(2 \cdot 3 - 5 \cdot 1)\)
   • \(= 1(25 - 21) - 2(10 - 7) + 3(6 - 5)\)
   • \(= 1(4) - 2(3) + 3(1) = 4 - 6 + 3 = 1\).

Ответ: The determinant is 1.

4.45. Ранг матрицы (Лекция 6, Пример 30)

Find the rank of the matrix: \(A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 2 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Use Gaussian Elimination: Reduce the matrix to row echelon form.
   • \(\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 2 \end{bmatrix} \xrightarrow[R_3 \to R_3-R_1]{R_2 \to R_2-2R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -1 \end{bmatrix}\).
2. Swap Rows: Swap \(R_2\) and \(R_3\) to get closer to echelon form.
   • \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\).
3. Count Non-Zero Rows: The matrix is now in row echelon form. There are two non-zero rows.

Ответ: The rank of the matrix is 2.

4.46. Обратная матрица через присоединённую (Лекция 6, Пример 31)

Find the inverse of the following matrices using the adjugate method: \(A = \begin{bmatrix} 2 & 1 & 1 \\ 4 & 3 & 3 \\ 6 & 7 & 12 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 4 & 1 & 8 \end{bmatrix}\)

Нажмите, чтобы увидеть решение

For Matrix A:

1. Calculate the determinant: \(\det(A) = 2(36-21) - 1(48-18) + 1(28-18) = 2(15) - 1(30) + 1(10) = 30-30+10 = 10\).
2. Find the matrix of cofactors: \(C = \begin{bmatrix} 15 & -30 & 10 \\ -5 & 18 & -8 \\ 0 & -2 & 2 \end{bmatrix}\).
3. Find the adjugate (transpose of cofactors): \(\text{adj}(A) = C^T = \begin{bmatrix} 15 & -5 & 0 \\ -30 & 18 & -2 \\ 10 & -8 & 2 \end{bmatrix}\).
4. Calculate the inverse: \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A) = \frac{1}{10}\begin{bmatrix} 15 & -5 & 0 \\ -30 & 18 & -2 \\ 10 & -8 & 2 \end{bmatrix}\).

For Matrix B:

1. Calculate the determinant: \(\det(B) = 1(8-3) - 0(...) + 2(2-4) = 5 - 4 = 1\).
2. Find the matrix of cofactors: \(C = \begin{bmatrix} 5 & -4 & -2 \\ 2 & 0 & -1 \\ -2 & 1 & 1 \end{bmatrix}\).
3. Find the adjugate: \(\text{adj}(B) = C^T = \begin{bmatrix} 5 & 2 & -2 \\ -4 & 0 & 1 \\ -2 & -1 & 1 \end{bmatrix}\).
4. Calculate the inverse: \(B^{-1} = \frac{1}{1}\begin{bmatrix} 5 & 2 & -2 \\ -4 & 0 & 1 \\ -2 & -1 & 1 \end{bmatrix}\).

Ответ: \(A^{-1} = \begin{bmatrix} 1.5 & -0.5 & 0 \\ -3 & 1.8 & -0.2 \\ 1 & -0.8 & 0.2 \end{bmatrix}\) \(B^{-1} = \begin{bmatrix} 5 & 2 & -2 \\ -4 & 0 & 1 \\ -2 & -1 & 1 \end{bmatrix}\)

4.47. Система: метод Гаусса (Лекция 6, Пример 32)

Use Gaussian elimination to solve the system: \[ \begin{cases} x + 2y + 3z = 6 \\ 2x + 4y + 7z = 13 \\ 3x + 7y + 11z = 20 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Write the Augmented Matrix:
   • \(\begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 2 & 4 & 7 & | & 13 \\ 3 & 7 & 11 & | & 20 \end{bmatrix}\)
2. Perform Row Operations:
   • \(R_2 \to R_2-2R_1\) and \(R_3 \to R_3-3R_1\)
   • \(\begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 0 & 0 & 1 & | & 1 \\ 0 & 1 & 2 & | & 2 \end{bmatrix}\)
3. Swap Rows: Swap \(R_2\) and \(R_3\).
   • \(\begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 0 & 1 & 2 & | & 2 \\ 0 & 0 & 1 & | & 1 \end{bmatrix}\)
4. Back Substitution:
   • \(z=1\).
   • \(y+2z = 2 \implies y+2(1)=2 \implies y=0\).
   • \(x+2y+3z = 6 \implies x+2(0)+3(1)=6 \implies x=3\).

Ответ: The solution is \(x=3, y=0, z=1\).

4.48. Определитель через свойства (Лекция 6, Пример 33)

Calculate the determinant using properties of determinants: \(\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}\).

Нажмите, чтобы увидеть решение

1. Apply Row Operations: The value of a determinant does not change when a multiple of one row is added to another. We will use this to create zeros.
   • \(R_2 \to R_2 - 4R_1\) and \(R_3 \to R_3 - 7R_1\).
   • \(\begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{vmatrix}\)
2. Analyze the Result: Notice that the third row is exactly 2 times the second row (\(R_3 = 2R_2\)).
3. Conclusion: If one row of a matrix is a scalar multiple of another row, the rows are linearly dependent, and the determinant of the matrix is 0.

Ответ: The determinant is 0.

4.49. Обратная матрица: Гаусс–Жордан (Лекция 6, Пример 34)

Find the inverse of the matrix using Gauss-Jordan elimination: \(B = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 5 & 2 \\ 1 & 3 & 4 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Set up the Augmented Matrix \([B|I]\):
   • \(\begin{bmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 2 & 5 & 2 & | & 0 & 1 & 0 \\ 1 & 3 & 4 & | & 0 & 0 & 1 \end{bmatrix}\)
2. Create Zeros Below the First Pivot:
   • \(R_2 \to R_2-2R_1\), \(R_3 \to R_3-R_1\)
   • \(\begin{bmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -2 & 1 & 0 \\ 0 & 1 & 3 & | & -1 & 0 & 1 \end{bmatrix}\)
3. Create Zero Below the Second Pivot:
   • \(R_3 \to R_3-R_2\)
   • \(\begin{bmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -2 & 1 & 0 \\ 0 & 0 & 3 & | & 1 & -1 & 1 \end{bmatrix}\)
4. Create Zeros Above the Pivots:
   • \(R_3 \to \frac{1}{3}R_3\): \(\begin{bmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -2 & 1 & 0 \\ 0 & 0 & 1 & | & 1/3 & -1/3 & 1/3 \end{bmatrix}\)
   • \(R_1 \to R_1-R_3\): \(\begin{bmatrix} 1 & 2 & 0 & | & 2/3 & 1/3 & -1/3 \\ 0 & 1 & 0 & | & -2 & 1 & 0 \\ 0 & 0 & 1 & | & 1/3 & -1/3 & 1/3 \end{bmatrix}\)
   • \(R_1 \to R_1-2R_2\): \(\begin{bmatrix} 1 & 0 & 0 & | & 14/3 & -5/3 & -1/3 \\ 0 & 1 & 0 & | & -2 & 1 & 0 \\ 0 & 0 & 1 & | & 1/3 & -1/3 & 1/3 \end{bmatrix}\)

Ответ: \(B^{-1} = \begin{bmatrix} 14/3 & -5/3 & -1/3 \\ -2 & 1 & 0 \\ 1/3 & -1/3 & 1/3 \end{bmatrix}\).

4.50. Параметр при вырожденной матрице (Лекция 6, Пример 35)

For what value(s) of \(k\) is the matrix singular? \(D = \begin{bmatrix} 1 & k & 1 \\ 2 & 5 & 2 \\ 1 & 3 & k \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Condition for Singularity: A matrix is singular if its determinant is zero.
2. Calculate the determinant in terms of \(k\):
   • \(\det(D) = 1(5k - 6) - k(2k - 2) + 1(6 - 5)\)
   • \(= 5k - 6 - 2k^2 + 2k + 1\)
   • \(= -2k^2 + 7k - 5\).
3. Set the determinant to zero and solve:
   • \(-2k^2 + 7k - 5 = 0\)
   • \(2k^2 - 7k + 5 = 0\)
   • \((2k - 5)(k - 1) = 0\).
4. Find the values of \(k\):
   • \(k = 1\) or \(k = 5/2\).

Ответ: The matrix is singular when \(k=1\) or \(k=5/2\).

4.51. Образ и ядро (столбцовое пространство и null space) (Лекция 6, Пример 36)

Determine the column space and null space of: \(C = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Find the Row Echelon Form:
   • \(\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} \xrightarrow[R_3 \to R_3-3R_1]{R_2 \to R_2-2R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\).
2. Determine the Column Space: The first column is a pivot column. The basis for the column space is the first column of the original matrix C.
   • Basis for Column Space: \(\left\{\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\right\}\).
3. Determine the Null Space: Solve \(C\mathbf{x} = \mathbf{0}\) using the RREF.
   • \(x_1 + 2x_2 + 3x_3 = 0\). \(x_1\) is the pivot variable, \(x_2, x_3\) are free.
   • \(x_1 = -2x_2 - 3x_3\).
   • The solution vector is \(\mathbf{x} = \begin{bmatrix} -2x_2-3x_3 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\).
   • Basis for Null Space: \(\left\{\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\right\}\).

Ответ: The column space is the line spanned by \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\). The null space is the plane spanned by \(\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\).

4.52. Проверка обратной матрицы (Лекция 6, Пример 37)

Verify that \(AA^{-1} = I\) for matrix \(A = \begin{bmatrix} 2 & 1 & 1 \\ 4 & 3 & 3 \\ 6 & 7 & 12 \end{bmatrix}\) using your result from question 12.

Нажмите, чтобы увидеть решение

1. Recall the Inverse: From question 12, \(A^{-1} = \frac{1}{10}\begin{bmatrix} 15 & -5 & 0 \\ -30 & 18 & -2 \\ 10 & -8 & 2 \end{bmatrix}\).
2. Perform the multiplication \(AA^{-1}\):
   • \(AA^{-1} = \frac{1}{10} \begin{bmatrix} 2 & 1 & 1 \\ 4 & 3 & 3 \\ 6 & 7 & 12 \end{bmatrix} \begin{bmatrix} 15 & -5 & 0 \\ -30 & 18 & -2 \\ 10 & -8 & 2 \end{bmatrix}\)
   • \(= \frac{1}{10} \begin{bmatrix} (30-30+10) & (-10+18-8) & (0-2+2) \\ (60-90+30) & (-20+54-24) & (0-6+6) \\ (90-210+120) & (-30+126-96) & (0-14+24) \end{bmatrix}\)
3. Simplify the result:
   • \(= \frac{1}{10} \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I\).

Ответ: The product \(AA^{-1}\) is the identity matrix \(I\), so the inverse is correct.

4.53. Система с параметром (Лекция 6, Пример 38)

For what values of \(k\) does the system have: (a) A unique solution? (b) Infinitely many solutions? (c) No solution? \[ \begin{cases} x + y + z = 1 \\ x + 2y + 4z = k \\ x + 4y + 10z = k^2 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Row Reduce the Augmented Matrix:
   • \(\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 4 & | & k \\ 1 & 4 & 10 & | & k^2 \end{bmatrix} \xrightarrow[R_3 \to R_3-R_1]{R_2 \to R_2-R_1} \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & k-1 \\ 0 & 3 & 9 & | & k^2-1 \end{bmatrix}\)
   • \(R_3 \to R_3-3R_2\): \(\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & k-1 \\ 0 & 0 & 0 & | & (k^2-1)-3(k-1) \end{bmatrix}\)
2. Analyze the Last Row:
   • The last row represents the equation \(0 = (k^2-1)-3(k-1)\).
   • \(0 = (k-1)(k+1) - 3(k-1) = (k-1)(k+1-3) = (k-1)(k-2)\).
3. Determine the Conditions:
   • (c) No solution: The system is inconsistent if the last row is \([0 \ 0 \ 0 \ | \ \text{non-zero}]\). This happens if \((k-1)(k-2) \neq 0\). So, no solution if \(k \neq 1\) and \(k \neq 2\).
   • (b) Infinitely many solutions: The system has infinite solutions if the last row is \([0 \ 0 \ 0 \ | \ 0]\). This happens if \((k-1)(k-2) = 0\). So, infinite solutions if \(k=1\) or \(k=2\).
   • (a) Unique solution: A unique solution requires a pivot in every variable column. Since the third column has no pivot (the last row is all zeros in the coefficient part), this system can never have a unique solution.

Ответ: (a) Never, (b) \(k=1\) or \(k=2\), (c) \(k \neq 1\) and \(k \neq 2\).

4.54. Система через обратную матрицу (Лекция 6, Пример 39)

Solve the system using matrix inversion: \[ \begin{cases} 2x + y + z = 7 \\ 4x + 3y + 3z = 17 \\ 6x + 7y + 12z = 44 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Write in Matrix Form: \(A\mathbf{x} = \mathbf{b}\), where \(A = \begin{bmatrix} 2 & 1 & 1 \\ 4 & 3 & 3 \\ 6 & 7 & 12 \end{bmatrix}\), \(\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(\mathbf{b} = \begin{bmatrix} 7 \\ 17 \\ 44 \end{bmatrix}\).
2. Use the Inverse: The solution is \(\mathbf{x} = A^{-1}\mathbf{b}\). The matrix A is the same as in question 12.
   • \(A^{-1} = \frac{1}{10}\begin{bmatrix} 15 & -5 & 0 \\ -30 & 18 & -2 \\ 10 & -8 & 2 \end{bmatrix}\).
3. Calculate the solution:
   • \(\mathbf{x} = \frac{1}{10}\begin{bmatrix} 15 & -5 & 0 \\ -30 & 18 & -2 \\ 10 & -8 & 2 \end{bmatrix} \begin{bmatrix} 7 \\ 17 \\ 44 \end{bmatrix}\)
   • \(= \frac{1}{10}\begin{bmatrix} 15(7) - 5(17) + 0(44) \\ -30(7) + 18(17) - 2(44) \\ 10(7) - 8(17) + 2(44) \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 105 - 85 \\ -210 + 306 - 88 \\ 70 - 136 + 88 \end{bmatrix}\)
4. Simplify:
   • \(= \frac{1}{10}\begin{bmatrix} 20 \\ 8 \\ 22 \end{bmatrix} = \begin{bmatrix} 2 \\ 0.8 \\ 2.2 \end{bmatrix}\).

Ответ: The solution is \(x=2, y=0.8, z=2.2\).

4.55. Линейная независимость и базис (Лекция 6, Пример 40)

Determine whether the following vectors are linearly independent and find a basis for their span: \(\vec{v_1}=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \vec{v_2}=\begin{bmatrix} 2 \\ 5 \\ 7 \end{bmatrix}, \vec{v_3}=\begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Check for Linear Independence: We can check if the determinant of the matrix formed by these vectors is non-zero.
   • \(\det(A) = \det \begin{bmatrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 7 & 5 \end{bmatrix}\)
   • \(= 1(25-21) - 2(10-9) + 1(14-15) = 4 - 2 - 1 = 1\).
2. Conclusion on Independence: Since the determinant is 1 (non-zero), the vectors are linearly independent.
3. Find a Basis for their Span: Since the three vectors are linearly independent in \(\mathbb{R}^3\), they already form a basis for their span. Their span is the entire space \(\mathbb{R}^3\).

Ответ: The vectors are linearly independent. A basis for their span is the set of vectors themselves: \(\left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} \right\}\).

4.56. Определитель произведения матриц (Лекция 6, Пример 41)

Find the determinant of the product: \(\begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 2 & 1 & 1 \end{vmatrix} \times \begin{vmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix}\).

Нажмите, чтобы увидеть решение

1. Use the Property \(\det(AB) = \det(A)\det(B)\): We can find the determinant of each matrix and then multiply the results.
2. Calculate Determinant of First Matrix (A):
   • \(\det(A) = 1(1-2) - 2(0-4) + 1(0-2) = -1 + 8 - 2 = 5\).
3. Calculate Determinant of Second Matrix (B):
   • \(\det(B) = 2(4-1) - 1(2-0) + 0(...) = 2(3) - 1(2) = 6 - 2 = 4\).
4. Multiply the Determinants:
   • \(\det(AB) = 5 \times 4 = 20\).

Ответ: The determinant of the product is 20.

4.57. Правило Крамера (Лекция 6, Пример 42)

Use Cramer’s rule to solve: \[ \begin{cases} 2x + y + z = 7 \\ x - y + 2z = 3 \\ 3x + 2y - z = 4 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Calculate the Main Determinant (D):
   • \(D = \det \begin{bmatrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{bmatrix} = 2(1-4) - 1(-1-6) + 1(2+3) = -6+7+5 = 6\).
2. Calculate \(D_x\):
   • \(D_x = \det \begin{bmatrix} 7 & 1 & 1 \\ 3 & -1 & 2 \\ 4 & 2 & -1 \end{bmatrix} = 7(1-4) - 1(-3-8) + 1(6+4) = -21+11+10 = 0\).
3. Calculate \(D_y\):
   • \(D_y = \det \begin{bmatrix} 2 & 7 & 1 \\ 1 & 3 & 2 \\ 3 & 4 & -1 \end{bmatrix} = 2(-3-8) - 7(-1-6) + 1(4-9) = -22+49-5 = 22\).
4. Calculate \(D_z\):
   • \(D_z = \det \begin{bmatrix} 2 & 1 & 7 \\ 1 & -1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = 2(-4-6) - 1(4-9) + 7(2+3) = -20+5+35 = 20\).
5. Find the Solution:
   • \(x = D_x/D = 0/6 = 0\).
   • \(y = D_y/D = 22/6 = 11/3\).
   • \(z = D_z/D = 20/6 = 10/3\).

Ответ: \(x=0, y=11/3, z=10/3\).

4.58. Доказательство \(\det(A^T) = \det(A)\) (Лекция 6, Пример 43)

Prove that for any \(3 \times 3\) matrix \(A\), \(\det(A^T) = \det(A)\).

Нажмите, чтобы увидеть решение

1. Define a general 3x3 matrix A:
   • \(A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\)
2. Calculate \(\det(A)\):
   • \(\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\).
3. Find the transpose \(A^T\):
   • \(A^T = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix}\)
4. Calculate \(\det(A^T)\):
   • \(\det(A^T) = a(ei - hf) - d(bi - hc) + g(bf - ec)\).
   • \(= aei - afh - bdi + cdh + bfg - ceg\).
5. Compare the expanded forms:
   • \(\det(A) = aei - afh - bdi + bfg + cdh - ceg\).
   • The terms are identical.

Ответ: By expanding the determinants of a general \(3 \times 3\) matrix and its transpose, we see that the resulting expressions are identical, thus proving \(\det(A^T) = \det(A)\).

4.59. Ранг и обратимость (Лекция 6, Пример 44)

Prove that if \(A\) is an \(n \times n\) matrix with rank \(n\), then \(A\) is invertible.

Нажмите, чтобы увидеть решение

1. Definition of Rank: An \(n \times n\) matrix has rank \(n\) if and only if its columns (and rows) are linearly independent.
2. Implication of Linear Independence: If the columns of \(A\) are linearly independent, the only solution to the homogeneous equation \(A\mathbf{x} = \mathbf{0}\) is the trivial solution \(\mathbf{x} = \mathbf{0}\).
3. Connection to Row Echelon Form: If the only solution to \(A\mathbf{x} = \mathbf{0}\) is the trivial one, it means that when we reduce the augmented matrix \([A|\mathbf{0}]\) to row echelon form, there are no free variables. For an \(n \times n\) matrix, this means there must be a pivot in every column.
4. Reduced Row Echelon Form: An \(n \times n\) matrix with a pivot in every column can be row-reduced to the \(n \times n\) identity matrix, \(I\).
5. Definition of Invertibility: A square matrix \(A\) is invertible if and only if its reduced row echelon form is the identity matrix. Since we have shown that \(\text{rank}(A)=n\) implies that the RREF of A is \(I\), it follows that A is invertible.

Ответ: A rank of \(n\) for an \(n \times n\) matrix implies its columns are linearly independent, which means its reduced row echelon form is the identity matrix, which is the definition of an invertible matrix.

4.60. Обратная к элементарной матрице (Лекция 6, Пример 45)

Find the inverse of the elementary matrix \(E = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) and verify that \(EE^{-1} = I\).

Нажмите, чтобы увидеть решение

1. Interpret the Matrix Operation: The matrix \(E\) represents the elementary row operation of multiplying the second row by 3.
2. Find the Inverse Operation: The inverse operation is to divide the second row by 3, or multiply it by 1/3.
3. Write the Inverse Matrix: The matrix that performs this inverse operation is:
   • \(E^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
4. Verify the Inverse: Multiply \(E\) by \(E^{-1}\).
   • \(EE^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
   • \(= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 \cdot (1/3) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I\).

Ответ: \(E^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\). The verification shows that \(EE^{-1} = I\).

4.61. Расстояние между скрещивающимися прямыми (Туториал 6, Пример 1)

Consider two skew lines \(L_1: \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\) and \(L_2: \frac{x-1}{2} = \frac{y}{-1} = \frac{z-2}{1}\). What is the smallest distance between these two lines?

Нажмите, чтобы увидеть решение

1. Identify Points and Direction Vectors:
   • For \(L_1\): A point on the line is \(P_1 = (0, 0, 0)\) and the direction vector is \(\vec{v_1} = (1, 2, 3)\).
   • For \(L_2\): A point on the line is \(P_2 = (1, 0, 2)\) and the direction vector is \(\vec{v_2} = (2, -1, 1)\).
2. Find a Common Normal Vector: The vector perpendicular to both lines is the cross product of their direction vectors.
   • \(\vec{n} = \vec{v_1} \times \vec{v_2} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{pmatrix}\)
   • \(= \mathbf{i}(2 - (-3)) - \mathbf{j}(1 - 6) + \mathbf{k}(-1 - 4) = 5\mathbf{i} + 5\mathbf{j} - 5\mathbf{k} = (5, 5, -5)\).
3. Find the Vector Connecting the Lines: Create a vector from a point on one line to a point on the other.
   • \(\vec{P_1P_2} = P_2 - P_1 = (1-0, 0-0, 2-0) = (1, 0, 2)\).
4. Use the Distance Formula: The distance between two skew lines is the absolute value of the scalar projection of the vector connecting them onto the common normal vector.
   • \(d = \frac{|\vec{P_1P_2} \cdot \vec{n}|}{||\vec{n}||}\).
5. Calculate:
   • \(\vec{P_1P_2} \cdot \vec{n} = (1)(5) + (0)(5) + (2)(-5) = 5 + 0 - 10 = -5\).
   • \(||\vec{n}|| = \sqrt{5^2 + 5^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}\).
   • \(d = \frac{|-5|}{5\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\).

Ответ: The smallest distance between the two lines is \(\frac{\sqrt{3}}{3}\).

4.62. Плоскость серединного перпендикуляра (Туториал 6, Пример 2)

Find the equation of the plane that is the perpendicular bisector of the segment joining the points \((3, 1, 0)\) and \((5, -1, 3)\).

Нажмите, чтобы увидеть решение

1. Find the Normal Vector: The normal vector to the plane is the vector connecting the two points.
   • \(\vec{n} = (5-3, -1-1, 3-0) = (2, -2, 3)\).
2. Find the Midpoint: The plane must pass through the midpoint of the segment.
   • \(M = (\frac{3+5}{2}, \frac{1+(-1)}{2}, \frac{0+3}{2}) = (4, 0, \frac{3}{2})\).
3. Use the Point-Normal Form: Use the normal vector \(\vec{n}=(2, -2, 3)\) and the midpoint \(M(4, 0, 3/2)\).
   • \(2(x - 4) - 2(y - 0) + 3(z - \frac{3}{2}) = 0\).
4. Simplify the Equation:
   • \(2x - 8 - 2y + 3z - \frac{9}{2} = 0\).
   • Multiply by 2 to clear the fraction: \(4x - 16 - 4y + 6z - 9 = 0\).
   • \(4x - 4y + 6z = 25\).

Ответ: The equation of the plane is \(4x - 4y + 6z = 25\).

4.63. Взаимное расположение прямой и плоскости (Туториал 6, Пример 3)

Consider the line \(L: \frac{x-1}{2} = \frac{y-3}{3} = -z\) and the plane \(\pi: 3x - 2y + 4z = -1\).

1. Find the point \(P_0\) of intersection.
2. Find an equation for the plane perpendicular to \(L\) that passes through \(P_0\).
3. Find parametric equations for the line that passes through \(P_0\) and is perpendicular to \(\pi\).

Нажмите, чтобы увидеть решение

1. Find the Point of Intersection \(P_0\):
   • First, write the line in parametric form: \(\frac{x-1}{2} = \frac{y-3}{3} = \frac{z}{-1} = t\).
   • \(x = 1+2t\), \(y = 3+3t\), \(z = -t\).
   • Substitute into the plane equation: \(3(1+2t) - 2(3+3t) + 4(-t) = -1\).
   • \(3 + 6t - 6 - 6t - 4t = -1 \implies -3 - 4t = -1 \implies -4t = 2 \implies t = -1/2\).
   • Substitute \(t\) back to find the point: \(x=1+2(-1/2)=0\), \(y=3+3(-1/2)=3/2\), \(z=-(-1/2)=1/2\).
   • \(P_0 = (0, 3/2, 1/2)\).
2. Find the Perpendicular Plane:
   • The direction vector of the line \(L\), \(\vec{v_L} = (2, 3, -1)\), is the normal vector for the new plane.
   • Using the point-normal form with \(P_0\): \(2(x-0) + 3(y-3/2) - 1(z-1/2) = 0\).
   • \(2x + 3y - 9/2 - z + 1/2 = 0 \implies 2x + 3y - z - 4 = 0\).
3. Find the Perpendicular Line:
   • The normal vector of the plane \(\pi\), \(\vec{n_\pi} = (3, -2, 4)\), is the direction vector for the new line.
   • The line passes through \(P_0 = (0, 3/2, 1/2)\).
   • The parametric equations are: \(x = 0+3s=3s\), \(y = 3/2-2s\), \(z = 1/2+4s\).

Ответ:

1. The point of intersection is \(P_0 = (0, 3/2, 1/2)\).
2. The equation of the perpendicular plane is \(2x + 3y - z = 4\).
3. The parametric equations of the perpendicular line are \(x=3s, y=3/2-2s, z=1/2+4s\).

4.64. Прямая пересечения двух плоскостей (Туториал 6, Пример 4)

Find the line of intersection of the planes given by \(x - z = 1\) and \(2x - 3y + 4z = 2\).

Нажмите, чтобы увидеть решение

1. Find the Direction Vector: The direction vector \(\vec{v}\) of the line is the cross product of the normal vectors of the planes.
   • \(\vec{n_1} = (1, 0, -1)\) and \(\vec{n_2} = (2, -3, 4)\).
   • \(\vec{v} = \vec{n_1} \times \vec{n_2} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ 2 & -3 & 4 \end{pmatrix} = (-3, -6, -3)\).
   • We can use the simpler parallel vector \(\vec{u} = (1, 2, 1)\).
2. Find a Point on the Line: Set one variable to 0 and solve the system. Let \(x=0\).
   • \(0 - z = 1 \implies z = -1\).
   • \(2(0) - 3y + 4(-1) = 2 \implies -3y - 4 = 2 \implies -3y = 6 \implies y = -2\).
   • A point on the line is \((0, -2, -1)\).
3. Write the Equation of the Line: Using the point and the direction vector \(\vec{u}\).
   • Parametric: \(x=t, y=-2+2t, z=-1+t\).

Ответ: The parametric equations for the line of intersection are \(x=t, y=-2+2t, z=-1+t\).

4.65. Взаимное расположение плоскостей (Туториал 6, Пример 5)

Which of the following planes are coincident, perpendicular, or intersecting?

1. \(x + 2y - 3z = 2\)
2. \(15x - 9y - z = 2\)
3. \(-2x - 4y + 6z = -4\)
4. \(5x - 3y - \frac{1}{2}z = 1\)

Нажмите, чтобы увидеть решение

Let’s find the normal vectors: \(\vec{n_1}=(1,2,-3)\), \(\vec{n_2}=(15,-9,-1)\), \(\vec{n_3}=(-2,-4,6)\), \(\vec{n_4}=(5,-3,-1/2)\).

1. Planes 1 and 3:
   • \(\vec{n_3} = -2\vec{n_1}\). The normal vectors are parallel.
   • Check the constant terms: The equation for plane 3 is \(-2(x+2y-3z) = -4\), which simplifies to \(x+2y-3z = 2\). This is the same as plane 1.
   • Result: Planes 1 and 3 are coincident.
2. Planes 2 and 4:
   • Multiply plane 4 by 3: \(15x - 9y - \frac{3}{2}z = 3\). The normal vector is parallel to \(\vec{n_2}\), but the constant term is different (\(3 \neq 2\)).
   • Result: Planes 2 and 4 are parallel (but not coincident).
3. Planes 1 and 2:
   • \(\vec{n_1} \cdot \vec{n_2} = (1)(15) + (2)(-9) + (-3)(-1) = 15 - 18 + 3 = 0\).
   • Result: Planes 1 and 2 are perpendicular. Since plane 3 is coincident with 1, planes 3 and 2 are also perpendicular.
4. Other pairs: All other pairs of planes have normal vectors that are neither parallel nor perpendicular, so they are simply intersecting. This includes pairs (1,4), (2,3), and (3,4).

Ответ:

• Coincident: (1, 3)
• Parallel: (2, 4)
• Perpendicular: (1, 2) and (2, 3)
• Intersecting: All other pairs.

4.66. Угол между прямой и плоскостью (Туториал 6, Пример 6)

What is the angle between the line \(x-3=y-2=1-z\) and the plane \(2\sqrt{3}x + \sqrt{2}y + \sqrt{2}z + 5 = 0\)?

Нажмите, чтобы увидеть решение

1. Identify Direction and Normal Vectors:
   • Line: \(\frac{x-3}{1} = \frac{y-2}{1} = \frac{z-1}{-1}\). Direction vector \(\vec{v} = (1, 1, -1)\).
   • Plane: Normal vector \(\vec{n} = (2\sqrt{3}, \sqrt{2}, \sqrt{2})\).
2. Use the Sine Formula: The angle \(\theta\) between a line and a plane is given by \(\sin(\theta) = \frac{|\vec{v} \cdot \vec{n}|}{||\vec{v}|| ||\vec{n}||}\).
3. Calculate:
   • \(\vec{v} \cdot \vec{n} = (1)(2\sqrt{3}) + (1)(\sqrt{2}) + (-1)(\sqrt{2}) = 2\sqrt{3}\).
   • \(||\vec{v}|| = \sqrt{1^2+1^2+(-1)^2} = \sqrt{3}\).
   • \(||\vec{n}|| = \sqrt{(2\sqrt{3})^2+(\sqrt{2})^2+(\sqrt{2})^2} = \sqrt{12+2+2} = \sqrt{16} = 4\).
   • \(\sin(\theta) = \frac{|2\sqrt{3}|}{(\sqrt{3})(4)} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2}\).
4. Find the angle:
   • \(\theta = \arcsin(1/2) = 30^\circ\).

Ответ: The angle between the line and the plane is \(30^\circ\).

4.67. Условие невырожденности матрицы (Туториал 6, Пример 7)

For which \(m\), the following matrix is nonsingular? \(A = \begin{bmatrix} -1 & 0 & m \\ m & 1 & 0 \\ 3m & 2 & -1 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Condition for Nonsingularity: A matrix is nonsingular if its determinant is non-zero.
2. Calculate the determinant:
   • \(\det(A) = -1(1(-1) - 0(2)) - 0(...) + m(m(2) - 1(3m))\)
   • \(= -1(-1) + m(2m - 3m) = 1 + m(-m) = 1 - m^2\).
3. Set the determinant to not equal zero:
   • \(1 - m^2 \neq 0 \implies m^2 \neq 1 \implies m \neq \pm 1\).

Ответ: The matrix is nonsingular for all \(m\) such that \(m \neq 1\) and \(m \neq -1\).

4.68. Условие единственности решения (Туториал 6, Пример 8)

For which value of \(m\), does the following linear system have a unique solution? \[ \begin{cases} x_1 + x_2 + x_3 = 2 \\ x_1 + mx_2 + x_3 = 1 \\ x_1 + x_2 + mx_3 = 3 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Condition for Unique Solution: A linear system has a unique solution if the determinant of the coefficient matrix is non-zero.
2. Calculate the determinant:
   • \(D = \det \begin{bmatrix} 1 & 1 & 1 \\ 1 & m & 1 \\ 1 & 1 & m \end{bmatrix}\)
   • \(= 1(m^2 - 1) - 1(m - 1) + 1(1 - m) = m^2 - 1 - m + 1 + 1 - m = m^2 - 2m + 1\).
3. Set the determinant to not equal zero:
   • \(m^2 - 2m + 1 \neq 0\).
   • \((m-1)^2 \neq 0 \implies m-1 \neq 0 \implies m \neq 1\).

Ответ: The system has a unique solution for all \(m \neq 1\).

4.69. Условие нетривиальных решений (Туториал 6, Пример 9)

Assume that the following linear system has nontrivial solutions. What can we say about \(m\)? \[ \begin{cases} x_1 + 2x_2 = 0 \\ -x_2 + mx_3 = 0 \\ 2x_1 + 2x_2 + 2x_3 = 0 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Condition for Nontrivial Solutions: A homogeneous system has nontrivial solutions if and only if the determinant of the coefficient matrix is zero.
2. Calculate the determinant:
   • \(D = \det \begin{bmatrix} 1 & 2 & 0 \\ 0 & -1 & m \\ 2 & 2 & 2 \end{bmatrix}\)
   • \(= 1(-2 - 2m) - 2(0 - 2m) + 0(...) = -2 - 2m + 4m = 2m - 2\).
3. Set the determinant to zero:
   • \(2m - 2 = 0 \implies 2m = 2 \implies m = 1\).

Ответ: For the system to have nontrivial solutions, \(m\) must be equal to 1.

4.70. Ранг и невырожденность (Туториал 6, Пример 10)

Consider the vectors \(\vec{a}=\begin{bmatrix}1\\2\\3\end{bmatrix}, \vec{b}=\begin{bmatrix}1\\-1\\2\end{bmatrix}, \vec{c}=\begin{bmatrix}-2\\-4\\-6\end{bmatrix}, \vec{d}=\begin{bmatrix}1\\2\\-3\end{bmatrix}, \vec{e}=\begin{bmatrix}3\\6\\9\end{bmatrix}\). Matrix A is a composition of these vectors.

1. Which columns should be excluded to make the matrix nonsingular?
2. Find the rank of A.

Нажмите, чтобы увидеть решение

1. Form the Matrix and Find Dependencies: Let \(A = [\vec{a} \ \vec{b} \ \vec{c} \ \vec{d} \ \vec{e}]\).
   • Notice that \(\vec{c} = -2\vec{a}\) and \(\vec{e} = 3\vec{a}\). Vectors \(\vec{c}\) and \(\vec{e}\) are linearly dependent on \(\vec{a}\).
   • To form a nonsingular matrix, we need to select a set of linearly independent columns. Since the vectors are in \(\mathbb{R}^3\), the maximum size of such a set is 3. We must exclude the dependent columns \(\vec{c}\) and \(\vec{e}\).
   • Let’s check if the remaining vectors \(\{\vec{a}, \vec{b}, \vec{d}\}\) are linearly independent by checking the determinant.
   • \(\det([\vec{a} \ \vec{b} \ \vec{d}]) = \det\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 2 \\ 3 & 2 & -3 \end{bmatrix} = 1(3-4)-1(-6-6)+1(4+3) = -1+12+7 = 18\).
   • Since the determinant is non-zero, \(\{\vec{a}, \vec{b}, \vec{d}\}\) is a linearly independent set.
2. Find the Rank of A:
   • The rank of A is the size of the largest set of linearly independent columns.
   • We have found a set of 3 linearly independent columns (\(\vec{a}, \vec{b}, \vec{d}\)). Since the vectors are in \(\mathbb{R}^3\), the rank cannot be greater than 3.

Ответ:

1. To make a nonsingular \(3 \times 3\) matrix, columns \(\vec{c}\) and \(\vec{e}\) must be excluded.
2. The rank of the full matrix A is 3.

4.71. RREF и исключение Гаусса (Туториал 6, Пример 11)

1. Find RREF of: \(\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 3 & 4 \end{bmatrix}\).
2. Solve using Gaussian elimination: \(2x+3y-z=1, x-y+2z=8, 3x+2y+z=9\).

Нажмите, чтобы увидеть решение

1. Find RREF:
   • \(\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 3 & 4 \end{bmatrix} \xrightarrow[R_3\to R_3-R_1]{R_2\to R_2-2R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \xrightarrow{R_3\to R_3-R_2} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\)
   • This is REF. To get RREF: \(R_1 \to R_1-2R_2 \implies \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\).
2. Solve System:
   • \(\begin{bmatrix} 2 & 3 & -1 & | & 1 \\ 1 & -1 & 2 & | & 8 \\ 3 & 2 & 1 & | & 9 \end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2} \begin{bmatrix} 1 & -1 & 2 & | & 8 \\ 2 & 3 & -1 & | & 1 \\ 3 & 2 & 1 & | & 9 \end{bmatrix}\)
   • \(\xrightarrow[R_3\to R_3-3R_1]{R_2\to R_2-2R_1} \begin{bmatrix} 1 & -1 & 2 & | & 8 \\ 0 & 5 & -5 & | & -15 \\ 0 & 5 & -5 & | & -15 \end{bmatrix} \xrightarrow{R_3\to R_3-R_2} \begin{bmatrix} 1 & -1 & 2 & | & 8 \\ 0 & 5 & -5 & | & -15 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}\)
   • From \(R_2\): \(5y-5z=-15 \implies y-z=-3 \implies y=z-3\). Let \(z=t\) (free variable). Then \(y=t-3\).
   • From \(R_1\): \(x-y+2z=8 \implies x-(t-3)+2t=8 \implies x-t+3+2t=8 \implies x+t=5 \implies x=5-t\).

Ответ:

1. RREF is \(\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\).
2. The solution is \(x=5-t, y=t-3, z=t\).

4.72. REF и RREF (Туториал 6, Пример 12)

1. Convert to REF: \(\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 & 2 & 3 & 4 \\ 3 & 6 & 9 & 12 \end{bmatrix}\).
2. Solve using RREF: \(x+y+z=6, 2x-y+z=3, x+2y-z=2\).

Нажмите, чтобы увидеть решение

1. Convert to REF:
   • Notice that \(R_2 = \frac{1}{2}R_1\) and \(R_3 = \frac{3}{2}R_1\). All rows are multiples of the first.
   • \(\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 & 2 & 3 & 4 \\ 3 & 6 & 9 & 12 \end{bmatrix} \xrightarrow[R_3\to R_3-3/2 R_1]{R_2\to R_2-1/2 R_1} \begin{bmatrix} 2 & 4 & 6 & 8 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\).
   • Or, simplifying the first row: \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\).
2. Solve using RREF:
   • \(\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & -1 & 1 & | & 3 \\ 1 & 2 & -1 & | & 2 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}\).
   • The solution is read directly: \(x=1, y=2, z=3\).

Ответ:

1. REF is \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\).
2. The solution is \(x=1, y=2, z=3\).

4.73. Анализ линейной системы (Туториал 6, Пример 13)

1. Find complete solution: \(x+2y-3z=4, 2x+4y-6z=8, 3x+6y-9z=12\).
2. Find RREF and rank of: \(\begin{bmatrix} 1 & -2 & 3 & -4 \\ 2 & -4 & 6 & -8 \\ 3 & -6 & 9 & -12 \end{bmatrix}\).
3. Solve homogeneous system: \(x+2y-z=0, 2x+4y-2z=0, 3x+6y-3z=0\).

Нажмите, чтобы увидеть решение

1. Complete Solution:
   • Notice all three equations are multiples of the first. The system reduces to \(x+2y-3z=4\).
   • Let \(y=s\) and \(z=t\) be free variables. Then \(x=4-2s+3t\).
   • Solution: \(\mathbf{x} = \begin{bmatrix} 4 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}\).
2. RREF and Rank:
   • All rows are multiples of the first.
   • RREF: \(\begin{bmatrix} 1 & -2 & 3 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\).
   • There is one non-zero row, so the rank is 1.
3. Solve Homogeneous System:
   • All equations are multiples of the first. The system is \(x+2y-z=0\).
   • Let \(y=s, z=t\). Then \(x=-2s+t\).
   • Solution: \(\mathbf{x} = s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\).

Ответ:

1. \(\mathbf{x} = (4-2s+3t, s, t)\).
2. RREF is as above, Rank = 1.
3. \(\mathbf{x} = (-2s+t, s, t)\).

4.74. Совместность системы (Туториал 6, Пример 14)

Determine consistency: \(x+y+z=1, 2x+2y+2z=2, 3x+3y+3z=4\).

Нажмите, чтобы увидеть решение

1. Analyze the Equations:
   • The second equation is \(2(x+y+z)=2\), which simplifies to \(x+y+z=1\). This is consistent with the first equation.
   • The third equation is \(3(x+y+z)=4\), which simplifies to \(x+y+z=4/3\).
2. Find the Contradiction: We have two conflicting requirements for the same expression: \(x+y+z=1\) and \(x+y+z=4/3\). Since \(1 \neq 4/3\), this is a contradiction.

Ответ: The system is inconsistent (has no solution).

4.75. Угол между векторами (Туториал 6, Пример 15)

If \(||\vec{v_1} \times \vec{v_2}|| = 1\) and \(\vec{v_1} \cdot \vec{v_2} = \sqrt{3}\), what is the angle between \(\vec{v_1}\) and \(\vec{v_2}\)?

Нажмите, чтобы увидеть решение

1. Use Geometric Definitions:
   • \(||\vec{v_1} \times \vec{v_2}|| = ||\vec{v_1}|| ||\vec{v_2}|| \sin(\theta) = 1\).
   • \(\vec{v_1} \cdot \vec{v_2} = ||\vec{v_1}|| ||\vec{v_2}|| \cos(\theta) = \sqrt{3}\).
2. Find the Tangent: Divide the first equation by the second.
   • \(\frac{||\vec{v_1}|| ||\vec{v_2}|| \sin(\theta)}{||\vec{v_1}|| ||\vec{v_2}|| \cos(\theta)} = \frac{1}{\sqrt{3}}\).
   • \(\tan(\theta) = \frac{1}{\sqrt{3}}\).
3. Solve for the Angle:
   • \(\theta = \arctan(\frac{1}{\sqrt{3}}) = 30^\circ\).

Ответ: The angle between the vectors is \(30^\circ\).

4.76. Вектор, ортогональный данным (Туториал 6, Пример 16)

If the vectors \(\vec{v_1}\) and \(\vec{v_2}\) are orthogonal and \(\vec{v_1} \times \vec{v_2} = (-6, 3, 1)\) and \(\vec{v_1} = (1, 1, 3)\), calculate vector \(\vec{v_2}\).

Нажмите, чтобы увидеть решение

1. Use Vector Triple Product: We can use the identity \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\).
2. Apply the Identity: Let \(\vec{a}=\vec{v_1}\), \(\vec{b}=\vec{v_1}\), \(\vec{c}=\vec{v_2}\).
   • \(\vec{v_1} \times (\vec{v_1} \times \vec{v_2}) = (\vec{v_1} \cdot \vec{v_2})\vec{v_1} - (\vec{v_1} \cdot \vec{v_1})\vec{v_2}\).
3. Use Given Information:
   • We are given that \(\vec{v_1}\) and \(\vec{v_2}\) are orthogonal, so \(\vec{v_1} \cdot \vec{v_2} = 0\).
   • The equation simplifies to \(\vec{v_1} \times (\vec{v_1} \times \vec{v_2}) = -||\vec{v_1}||^2 \vec{v_2}\).
4. Solve for \(\vec{v_2}\):
   • \(\vec{v_2} = -\frac{1}{||\vec{v_1}||^2} [\vec{v_1} \times (\vec{v_1} \times \vec{v_2})]\).
5. Calculate:
   • \(||\vec{v_1}||^2 = 1^2+1^2+3^2 = 11\).
   • \(\vec{v_1} \times (\vec{v_1} \times \vec{v_2}) = (1,1,3) \times (-6,3,1) = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 3 \\ -6 & 3 & 1 \end{pmatrix} = (1-9, -18-1, 3+6) = (-8, -19, 9)\).
   • \(\vec{v_2} = -\frac{1}{11}(-8, -19, 9) = (\frac{8}{11}, \frac{19}{11}, -\frac{9}{11})\).

Ответ: \(\vec{v_2} = (\frac{8}{11}, \frac{19}{11}, -\frac{9}{11})\).

4.77. Скалярное произведение векторов (Туториал 6, Пример 17)

If \(\vec{a}\) and \(\vec{b}\) are vectors with properties \(||\vec{a}|| = ||\vec{b}|| = 2\) and \(||\vec{a} - \vec{b}|| = 2\sqrt{3}\), then what is \((\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b})\)?

Нажмите, чтобы увидеть решение

1. Find \(\vec{a} \cdot \vec{b}\):
   • \(||\vec{a} - \vec{b}||^2 = (2\sqrt{3})^2 = 12\).
   • \(||\vec{a} - \vec{b}||^2 = (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) = ||\vec{a}||^2 - 2(\vec{a}\cdot\vec{b}) + ||\vec{b}||^2\).
   • \(12 = 2^2 - 2(\vec{a}\cdot\vec{b}) + 2^2 = 8 - 2(\vec{a}\cdot\vec{b})\).
   • \(4 = -2(\vec{a}\cdot\vec{b}) \implies \vec{a}\cdot\vec{b} = -2\).
2. Expand the Target Expression:
   • \((\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \vec{a}\cdot\vec{a} - 2(\vec{a}\cdot\vec{b}) + \vec{b}\cdot\vec{a} - 2(\vec{b}\cdot\vec{b})\)
   • \(= ||\vec{a}||^2 - \vec{a}\cdot\vec{b} - 2||\vec{b}||^2\).
3. Substitute the Known Values:
   • \(= 2^2 - (-2) - 2(2^2) = 4 + 2 - 8 = -2\).

Ответ: The result is -2.

4.78. Параметрические уравнения прямой (Туториал 6, Задание 1)

Find the parametric equations for the line through the points \((-1, 1, 0)\) and \((-2, 5, 7)\).

Нажмите, чтобы увидеть решение

1. Find the Direction Vector: The direction vector \(\vec{v}\) of the line is the vector from one point to the other.
   • \(\vec{v} = (-2 - (-1), 5 - 1, 7 - 0) = (-1, 4, 7)\).
2. Choose a Point on the Line: We can use either of the given points. Let’s choose \(P_0 = (-1, 1, 0)\).
3. Write the Parametric Equations: The parametric equations of a line are given by \(x = x_0 + at\), \(y = y_0 + bt\), \(z = z_0 + ct\), where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is the direction vector.
   • \(x = -1 + (-1)t = -1 - t\)
   • \(y = 1 + 4t\)
   • \(z = 0 + 7t = 7t\)

Ответ: The parametric equations are \(x = -1 - t\), \(y = 1 + 4t\), \(z = 7t\).

4.79. Расстояние от начала координат до прямой (Туториал 6, Задание 2)

Find the distance from the origin to the line through the point \((-3, -3, 3)\) parallel to the vector \(\langle 4, -2, -4 \rangle\).

Нажмите, чтобы увидеть решение

1. Identify the Given Information:
   • The point to find the distance from is \(P = (0, 0, 0)\).
   • A point on the line is \(Q = (-3, -3, 3)\).
   • The direction vector of the line is \(\vec{v} = (4, -2, -4)\). We can use a simpler, parallel vector by dividing by 2: \(\vec{u} = (2, -1, -2)\).
2. Use the Distance Formula: The distance from a point \(P\) to a line is \(d = \frac{||\vec{QP} \times \vec{u}||}{||\vec{u}||}\).
3. Find the Vector \(\vec{QP}\):
   • \(\vec{QP} = P - Q = (0 - (-3), 0 - (-3), 0 - 3) = (3, 3, -3)\).
4. Calculate the Cross Product \(\vec{QP} \times \vec{u}\):
   • \(\vec{QP} \times \vec{u} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 3 & -3 \\ 2 & -1 & -2 \end{pmatrix}\)
   • \(= \mathbf{i}(3(-2) - (-3)(-1)) - \mathbf{j}(3(-2) - (-3)(2)) + \mathbf{k}(3(-1) - 3(2))\)
   • \(= \mathbf{i}(-6 - 3) - \mathbf{j}(-6 + 6) + \mathbf{k}(-3 - 6) = -9\mathbf{i} - 9\mathbf{k} = (-9, 0, -9)\).
5. Calculate the Magnitudes:
   • \(||\vec{QP} \times \vec{u}|| = \sqrt{(-9)^2 + 0^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}\).
   • \(||\vec{u}|| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\).
6. Calculate the Distance:
   • \(d = \frac{9\sqrt{2}}{3} = 3\sqrt{2}\).

Ответ: The distance from the origin to the line is \(3\sqrt{2}\).

4.80. Расстояние между двумя прямыми (Туториал 6, Задание 3)

Determine the distance between the two lines given by: \(L_1: \frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 2}{-2}\) \(L_2: \frac{x}{2} = \frac{y - 2}{-1} = \frac{z - 3}{-2}\)

Нажмите, чтобы увидеть решение

1. Check for Parallelism:
   • Direction vector of \(L_1\): \(\vec{v_1} = (2, -1, -2)\).
   • Direction vector of \(L_2\): \(\vec{v_2} = (2, -1, -2)\).
   • Since \(\vec{v_1} = \vec{v_2}\), the lines are parallel.
2. Use the Distance Formula for Parallel Lines: The distance between two parallel lines is \(d = \frac{||\vec{P_1 P_2} \times \vec{v}||}{||\vec{v}||}\), where \(P_1\) is a point on the first line, \(P_2\) is a point on the second line, and \(\vec{v}\) is the common direction vector.
3. Identify Points and the Direction Vector:
   • From \(L_1\), a point is \(P_1 = (1, -1, 2)\).
   • From \(L_2\), a point is \(P_2 = (0, 2, 3)\).
   • The direction vector is \(\vec{v} = (2, -1, -2)\).
4. Find the Vector \(\vec{P_1 P_2}\):
   • \(\vec{P_1 P_2} = P_2 - P_1 = (0-1, 2-(-1), 3-2) = (-1, 3, 1)\).
5. Calculate the Cross Product \(\vec{P_1 P_2} \times \vec{v}\):
   • \(\vec{P_1 P_2} \times \vec{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 1 \\ 2 & -1 & -2 \end{pmatrix}\)
   • \(= \mathbf{i}(3(-2) - 1(-1)) - \mathbf{j}((-1)(-2) - 1(2)) + \mathbf{k}((-1)(-1) - 3(2))\)
   • \(= \mathbf{i}(-6 + 1) - \mathbf{j}(2 - 2) + \mathbf{k}(1 - 6) = -5\mathbf{i} - 5\mathbf{k} = (-5, 0, -5)\).
6. Calculate the Magnitudes:
   • \(||\vec{P_1 P_2} \times \vec{v}|| = \sqrt{(-5)^2 + 0^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\).
   • \(||\vec{v}|| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\).
7. Calculate the Distance:
   • \(d = \frac{5\sqrt{2}}{3}\).

Ответ: The distance between the two lines is \(\frac{5\sqrt{2}}{3}\).

4.81. Плоскость через три точки (Туториал 6, Задание 4)

Find the equation of the plane through the points \(A(1, 0, 2)\), \(B(-1, 3, 4)\), and \(C(3, 5, 7)\).

Нажмите, чтобы увидеть решение

1. Find Two Vectors in the Plane: Create two vectors from the given points.
   • \(\vec{AB} = B - A = (-1-1, 3-0, 4-2) = (-2, 3, 2)\).
   • \(\vec{AC} = C - A = (3-1, 5-0, 7-2) = (2, 5, 5)\).
2. Find the Normal Vector: The normal vector \(\vec{n}\) to the plane is perpendicular to both \(\vec{AB}\) and \(\vec{AC}\). We find it by taking their cross product.
   • \(\vec{n} = \vec{AB} \times \vec{AC} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 3 & 2 \\ 2 & 5 & 5 \end{pmatrix}\)
   • \(= \mathbf{i}(3\cdot5 - 2\cdot5) - \mathbf{j}((-2)\cdot5 - 2\cdot2) + \mathbf{k}((-2)\cdot5 - 3\cdot2)\)
   • \(= \mathbf{i}(15-10) - \mathbf{j}(-10-4) + \mathbf{k}(-10-6) = 5\mathbf{i} + 14\mathbf{j} - 16\mathbf{k} = (5, 14, -16)\).
3. Use the Point-Normal Form: Use the normal vector \(\vec{n}=(5, 14, -16)\) and one of the points, say \(A(1, 0, 2)\).
   • \(5(x - 1) + 14(y - 0) - 16(z - 2) = 0\).
4. Simplify to the General Form:
   • \(5x - 5 + 14y - 16z + 32 = 0\).
   • \(5x + 14y - 16z + 27 = 0\).

Ответ: The equation of the plane is \(5x + 14y - 16z = -27\).

4.82. Плоскость через точку и прямую (Туториал 6, Задание 5)

Determine the equation of the plane that contains the point \((1, -1, 2)\) and the line \(2x = y = \frac{5z}{2}\).

Нажмите, чтобы увидеть решение

1. Find the Direction Vector and a Point on the Line:
   • Rewrite the line’s equation in symmetric form: \(\frac{x}{1/2} = \frac{y}{1} = \frac{z}{2/5}\).
   • The direction vector of the line is \(\vec{v} = (1/2, 1, 2/5)\). To avoid fractions, we can use a parallel vector \(\vec{v'} = 10\vec{v} = (5, 10, 4)\).
   • A point on the line is found by setting the expressions to 0, which gives \(Q = (0, 0, 0)\).
2. Find Two Vectors in the Plane:
   • One vector in the plane is the direction vector of the line, \(\vec{v'} = (5, 10, 4)\).
   • Another vector in the plane is the vector from the point on the line \(Q(0,0,0)\) to the given point \(P(1,-1,2)\). Let’s call this \(\vec{u} = P - Q = (1, -1, 2)\).
3. Find the Normal Vector: The normal vector \(\vec{n}\) is the cross product of the two vectors in the plane.
   • \(\vec{n} = \vec{u} \times \vec{v'} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 5 & 10 & 4 \end{pmatrix}\)
   • \(= \mathbf{i}(-1\cdot4 - 2\cdot10) - \mathbf{j}(1\cdot4 - 2\cdot5) + \mathbf{k}(1\cdot10 - (-1)\cdot5)\)
   • \(= \mathbf{i}(-4-20) - \mathbf{j}(4-10) + \mathbf{k}(10+5) = -24\mathbf{i} + 6\mathbf{j} + 15\mathbf{k} = (-24, 6, 15)\).
4. Use the Point-Normal Form: Use the normal vector \(\vec{n}=(-24, 6, 15)\) and a point in the plane, like the origin \(Q(0,0,0)\).
   • \(-24(x - 0) + 6(y - 0) + 15(z - 0) = 0\).
   • \(-24x + 6y + 15z = 0\). We can divide by -3 for a simpler form.
   • \(8x - 2y - 5z = 0\).

Ответ: The equation of the plane is \(8x - 2y - 5z = 0\).

4.83. Компланарны ли четыре точки? (Туториал 6, Задание 6)

Are the four points \((2, 3, 2)\), \((1, -1, -3)\), \((1, 0, -1)\), and \((5, 9, 5)\) coplanar? Why?

Нажмите, чтобы увидеть решение

1. Form Three Vectors: To check for coplanarity, we can form three vectors from the four points and see if they are coplanar. Let the points be \(P_1, P_2, P_3, P_4\). Let’s create vectors starting from \(P_2\):
   • \(\vec{P_2 P_1} = (2-1, 3-(-1), 2-(-3)) = (1, 4, 5)\).
   • \(\vec{P_2 P_3} = (1-1, 0-(-1), -1-(-3)) = (0, 1, 2)\).
   • \(\vec{P_2 P_4} = (5-1, 9-(-1), 5-(-3)) = (4, 10, 8)\).
2. Use the Scalar Triple Product: Three vectors are coplanar if the volume of the parallelepiped they form is zero. This volume is given by the absolute value of their scalar triple product.
3. Calculate the Scalar Triple Product (Determinant):
   • Volume \(= \left| \det \begin{pmatrix} 1 & 4 & 5 \\ 0 & 1 & 2 \\ 4 & 10 & 8 \end{pmatrix} \right|\)
   • \(= |1(1\cdot8 - 2\cdot10) - 4(0\cdot8 - 2\cdot4) + 5(0\cdot10 - 1\cdot4)|\)
   • \(= |1(8-20) - 4(0-8) + 5(0-4)|\)
   • \(= |1(-12) - 4(-8) + 5(-4)| = |-12 + 32 - 20| = |0|\).
4. Conclusion: Since the scalar triple product is zero, the three vectors lie on the same plane. Therefore, the four original points are coplanar.

Ответ: Yes, the four points are coplanar because the scalar triple product of the three vectors formed by them is zero.

4.84. Площадь треугольника и объём параллелепипеда (Повтор перед контрольной, N1)

Given the points \(A(1, 2, 3)\), \(B(2, -1, 1)\), \(C(3, 1, -2)\), and \(D(0, 1, 2)\), find the area of triangle \(ABC\) and the volume of the parallelepiped formed by the vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\).

Нажмите, чтобы увидеть решение

1. Define the vectors originating from point A:
   • \(\vec{AB} = B - A = (2-1, -1-2, 1-3) = (1, -3, -2)\).
   • \(\vec{AC} = C - A = (3-1, 1-2, -2-3) = (2, -1, -5)\).
   • \(\vec{AD} = D - A = (0-1, 1-2, 2-3) = (-1, -1, -1)\).
2. Calculate the Area of Triangle ABC:
   • The area is half the magnitude of the cross product of two side vectors: Area \(= \frac{1}{2}||\vec{AB} \times \vec{AC}||\).
   • First, compute the cross product: \(\vec{AB} \times \vec{AC} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & -2 \\ 2 & -1 & -5 \end{pmatrix} = \mathbf{i}(15 - 2) - \mathbf{j}(-5 - (-4)) + \mathbf{k}(-1 - (-6)) = 13\mathbf{i} + \mathbf{j} + 5\mathbf{k}\).
   • Next, find the magnitude: \(||\vec{AB} \times \vec{AC}|| = \sqrt{13^2 + 1^2 + 5^2} = \sqrt{169 + 1 + 25} = \sqrt{195}\).
   • The area is \(\frac{\sqrt{195}}{2}\).
3. Calculate the Volume of the Parallelepiped:
   • The volume is the absolute value of the scalar triple product: \(V = |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}|\).
   • Using the cross product from the previous step: \(V = |(13, 1, 5) \cdot (-1, -1, -1)|\).
   • \(V = |(13)(-1) + (1)(-1) + (5)(-1)| = |-13 - 1 - 5| = |-19| = 19\).

Ответ:

• The area of triangle ABC is \(\frac{\sqrt{195}}{2}\).
• The volume of the parallelepiped is \(19\).

4.85. Линейная независимость векторов (Повтор перед контрольной, N2)

Determine if the vectors \(\vec{a} = (1, 2, 3)\), \(\vec{b} = (2, 5, 7)\), and \(\vec{c} = (1, 3, 5)\) are linearly independent.

Нажмите, чтобы увидеть решение

1. Condition for Linear Independence: Three vectors in \(\mathbb{R}^3\) are linearly independent if and only if the determinant of the matrix formed by them is non-zero.
2. Form the Matrix: Create a matrix where the vectors are the rows (or columns).
   • \(M = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 3 & 5 \end{bmatrix}\).
3. Calculate the Determinant:
   • \(\det(M) = 1(5\cdot5 - 7\cdot3) - 2(2\cdot5 - 7\cdot1) + 3(2\cdot3 - 5\cdot1)\)
   • \(= 1(25-21) - 2(10-7) + 3(6-5)\)
   • \(= 1(4) - 2(3) + 3(1) = 4 - 6 + 3 = 1\).
4. Conclusion: Since the determinant is \(1\) (which is non-zero), the vectors are linearly independent.

Ответ: Yes, the vectors are linearly independent.

4.86. Расстояние от точки до прямой (Повтор перед контрольной, N3)

Find the distance from the point \(P(2, -1, 3)\) to the line passing through points \(A(1, 0, 1)\) and \(B(3, 2, 0)\).

Нажмите, чтобы увидеть решение

1. Find the Direction Vector of the Line:
   • \(\vec{v} = \vec{AB} = B - A = (3-1, 2-0, 0-1) = (2, 2, -1)\).
2. Find a Vector from the Line to the Point:
   • \(\vec{u} = \vec{AP} = P - A = (2-1, -1-0, 3-1) = (1, -1, 2)\).
3. Use the Distance Formula: The distance is given by \(d = \frac{||\vec{u} \times \vec{v}||}{||\vec{v}||}\).
4. Calculate the Cross Product:
   • \(\vec{u} \times \vec{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 2 & 2 & -1 \end{pmatrix} = \mathbf{i}(1-4) - \mathbf{j}(-1-4) + \mathbf{k}(2-(-2)) = (-3, 5, 4)\).
5. Calculate the Magnitudes:
   • \(||\vec{u} \times \vec{v}|| = \sqrt{(-3)^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}\).
   • \(||\vec{v}|| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\).
6. Calculate the Distance:
   • \(d = \frac{5\sqrt{2}}{3}\).

Ответ: The distance is \(\frac{5\sqrt{2}}{3}\).

4.87. Обратная матрица через присоединённую (Повтор перед контрольной, N4)

Given the matrix \(M = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 4 & 1 & 8 \end{bmatrix}\), find its inverse using the adjugate method.

Нажмите, чтобы увидеть решение

1. Use the Formula: The inverse is given by \(M^{-1} = \frac{1}{\det(M)}\text{adj}(M)\).
2. Calculate the Determinant:
   • \(\det(M) = 1(1\cdot8 - 3\cdot1) - 0(...) + 2(2\cdot1 - 1\cdot4) = 1(5) + 2(-2) = 1\).
3. Find the Matrix of Cofactors (\(M_c\)):
   • \(C_{11} = (8-3)=5\), \(C_{12} = -(16-12)=-4\), \(C_{13} = (2-4)=-2\).
   • \(C_{21} = -(0-2)=2\), \(C_{22} = (8-8)=0\), \(C_{23} = -(1-0)=-1\).
   • \(C_{31} = (0-2)=-2\), \(C_{32} = -(3-4)=1\), \(C_{33} = (1-0)=1\).
   • \(M_c = \begin{bmatrix} 5 & -4 & -2 \\ 2 & 0 & -1 \\ -2 & 1 & 1 \end{bmatrix}\).
4. Find the Adjugate Matrix: The adjugate is the transpose of the cofactor matrix, \(\text{adj}(M) = M_c^T\).
   • \(\text{adj}(M) = \begin{bmatrix} 5 & 2 & -2 \\ -4 & 0 & 1 \\ -2 & -1 & 1 \end{bmatrix}\).
5. Calculate the Inverse:
   • \(M^{-1} = \frac{1}{1} \begin{bmatrix} 5 & 2 & -2 \\ -4 & 0 & 1 \\ -2 & -1 & 1 \end{bmatrix}\).

Ответ: \(M^{-1} = \begin{bmatrix} 5 & 2 & -2 \\ -4 & 0 & 1 \\ -2 & -1 & 1 \end{bmatrix}\).

4.88. Параметр при невырожденной матрице (Повтор перед контрольной, N5)

For what values of \(k\) is the matrix \(D = \begin{bmatrix} 1 & k & 1 \\ 2 & 5 & 2 \\ 1 & 3 & k \end{bmatrix}\) nonsingular?

Нажмите, чтобы увидеть решение

1. Condition for Nonsingularity: A matrix is nonsingular if and only if its determinant is non-zero.
2. Calculate the Determinant:
   • \(\det(D) = 1(5k - 6) - k(2k - 2) + 1(6 - 5)\)
   • \(= 5k - 6 - 2k^2 + 2k + 1 = -2k^2 + 7k - 5\).
3. Find Values for Singularity: First, find the values of \(k\) that make the matrix singular by setting the determinant to zero.
   • \(-2k^2 + 7k - 5 = 0 \implies 2k^2 - 7k + 5 = 0\).
4. Solve the Quadratic Equation:
   • Using the quadratic formula: \(k = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(5)}}{2(2)} = \frac{7 \pm \sqrt{49 - 40}}{4} = \frac{7 \pm \sqrt{9}}{4} = \frac{7 \pm 3}{4}\).
   • The solutions are \(k_1 = \frac{7+3}{4} = \frac{10}{4} = \frac{5}{2}\) and \(k_2 = \frac{7-3}{4} = \frac{4}{4} = 1\).
5. Conclusion: The matrix is singular for \(k=1\) and \(k=5/2\). Therefore, it is nonsingular for all other values.

Ответ: The matrix is nonsingular for all real numbers \(k\) except \(k=1\) and \(k=5/2\).

4.89. Система линейных уравнений (Повтор перед контрольной, N6)

Solve the system of linear equations: \[ \begin{cases} 2x + y + z = 7 \\ 4x + 3y + 3z = 17 \\ 6x + 7y + 12z = 44 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Write the Augmented Matrix:
   • \(\begin{bmatrix} 2 & 1 & 1 & | & 7 \\ 4 & 3 & 3 & | & 17 \\ 6 & 7 & 12 & | & 44 \end{bmatrix}\)
2. Use Gaussian Elimination:
   • \(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 3R_1\): \(\begin{bmatrix} 2 & 1 & 1 & | & 7 \\ 0 & 1 & 1 & | & 3 \\ 0 & 4 & 9 & | & 23 \end{bmatrix}\)
   • \(R_3 \to R_3 - 4R_2\): \(\begin{bmatrix} 2 & 1 & 1 & | & 7 \\ 0 & 1 & 1 & | & 3 \\ 0 & 0 & 5 & | & 11 \end{bmatrix}\)
3. Perform Back Substitution:
   • From the third row: \(5z = 11 \implies z = 11/5\).
   • From the second row: \(y + z = 3 \implies y = 3 - 11/5 = 4/5\).
   • From the first row: \(2x + y + z = 7 \implies 2x + 4/5 + 11/5 = 7 \implies 2x + 15/5 = 7 \implies 2x + 3 = 7 \implies 2x = 4 \implies x = 2\).

Ответ: The solution is \((x, y, z) = (2, 4/5, 11/5)\).

4.90. Канонические уравнения перпендикулярной прямой (Повтор перед контрольной, N7)

Find the symmetric equation of the line passing through point \(P(1, -2, 3)\) and perpendicular to the plane \(\alpha: 2x+y-z=4\).

Нажмите, чтобы увидеть решение

1. Identify the Direction Vector: The direction vector of a line perpendicular to a plane is the plane’s normal vector.
   • The normal vector of the plane is \(\vec{n} = (2, 1, -1)\).
2. Use the Point and Direction Vector:
   • Point on the line: \(P_0 = (1, -2, 3)\).
   • Direction vector: \(\vec{v} = (2, 1, -1)\).
3. Write the Symmetric Equation: The formula is \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).
   • \(\frac{x-1}{2} = \frac{y-(-2)}{1} = \frac{z-3}{-1}\).

Ответ: The symmetric equation of the line is \(\frac{x-1}{2} = \frac{y+2}{1} = \frac{z-3}{-1}\).

4.91. Система: правило Крамера (Повтор перед контрольной, N8)

Solve the system of linear equations using Cramer’s rule: \[ \begin{cases} 2x + y + z = 7 \\ x - y + 2z = 3 \\ 3x + 2y - z = 4 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Calculate the Main Determinant (D):
   • \(D = \det \begin{bmatrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{bmatrix} = 2(1-4) - 1(-1-6) + 1(2 - (-3)) = -6 + 7 + 5 = 6\).
2. Calculate the Determinant for x (\(D_x\)):
   • \(D_x = \det \begin{bmatrix} 7 & 1 & 1 \\ 3 & -1 & 2 \\ 4 & 2 & -1 \end{bmatrix} = 7(1-4) - 1(-3-8) + 1(6 - (-4)) = -21 + 11 + 10 = 0\).
3. Calculate the Determinant for y (\(D_y\)):
   • \(D_y = \det \begin{bmatrix} 2 & 7 & 1 \\ 1 & 3 & 2 \\ 3 & 4 & -1 \end{bmatrix} = 2(-3-8) - 7(-1-6) + 1(4-9) = -22 + 49 - 5 = 22\).
4. Calculate the Determinant for z (\(D_z\)):
   • \(D_z = \det \begin{bmatrix} 2 & 1 & 7 \\ 1 & -1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = 2(-4-6) - 1(4-9) + 7(2 - (-3)) = -20 + 5 + 35 = 20\).
5. Find the Solution:
   • \(x = D_x/D = 0/6 = 0\).
   • \(y = D_y/D = 22/6 = 11/3\).
   • \(z = D_z/D = 20/6 = 10/3\).

Ответ: The solution is \((x, y, z) = (0, 11/3, 10/3)\).

4.92. Ядро и образ (null space и column space) (Повтор перед контрольной, N9)

Find the null space and column space of the matrix \(M = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

1. Find the Row Echelon Form:
   • Perform row operations: \(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 3R_1\).
   • \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\).
2. Find the Null Space: Solve the system \(M\mathbf{x} = \mathbf{0}\), which simplifies to the single equation \(x_1 + 2x_2 + 3x_3 = 0\).
   • The first variable, \(x_1\), is the pivot variable. The other two, \(x_2\) and \(x_3\), are free variables. Let \(x_2 = \alpha\) and \(x_3 = \beta\).
   • \(x_1 = -2\alpha - 3\beta\).
   • The general solution is \(\mathbf{x} = \begin{bmatrix} -2\alpha - 3\beta \\ \alpha \\ \beta \end{bmatrix} = \alpha\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + \beta\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\).
   • The null space is the span of these two vectors.
3. Find the Column Space: The column space is spanned by the pivot columns of the original matrix.
   • The first column is the only pivot column.
   • The column space is the span of the first column of M.

Ответ:

• Null Space: The plane spanned by the vectors \(\left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\}\).
• Column Space: The line spanned by the vector \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\).

4.93. Система с параметром (Повтор перед контрольной, N10)

For what values of \(k\) does the system have a unique solution, infinitely many solutions, or no solution? \[ \begin{cases} x + y + z = 1 \\ x + 2y + 4z = k \\ x + 4y + 10z = k^2 \end{cases} \]

Нажмите, чтобы увидеть решение

1. Row Reduce the Augmented Matrix:
   • \(\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 4 & | & k \\ 1 & 4 & 10 & | & k^2 \end{bmatrix}\)
   • \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\): \(\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & k-1 \\ 0 & 3 & 9 & | & k^2-1 \end{bmatrix}\)
   • \(R_3 \to R_3 - 3R_2\): \(\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & k-1 \\ 0 & 0 & 0 & | & (k^2-1) - 3(k-1) \end{bmatrix}\)
2. Analyze the Last Row: The equation for the last row is \(0 = (k^2-1) - 3(k-1)\).
   • Factor the expression: \(0 = (k-1)(k+1) - 3(k-1) = (k-1)(k+1-3) = (k-1)(k-2)\).
3. Determine the Conditions:
   • No solution: This occurs when the last row is of the form \([0, 0, 0 | c]\) where \(c \neq 0\). This happens when \((k-1)(k-2) \neq 0\), i.e., \(k \neq 1\) and \(k \neq 2\).
   • Infinitely many solutions: This occurs when the last row is \([0, 0, 0 | 0]\). This happens when \((k-1)(k-2) = 0\), i.e., \(k=1\) or \(k=2\).
   • Unique solution: This would require a pivot in every column of the coefficient matrix. Since the third column has no pivot, a unique solution is impossible for any value of \(k\).

Ответ:

• Unique solution: Never
• Infinitely many solutions: \(k=1\) or \(k=2\)
• No solution: \(k \neq 1\) and \(k \neq 2\)

4.94. Пересечение прямой и плоскости (Контрольная, Задание 1a)

Find intersection point \(P_0\) between line \(L: \frac{x-1}{2} = \frac{y-3}{3} = -z\) and plane \(\pi: 3x - 2y + 4z = -1\).

Нажмите, чтобы увидеть решение

Key Concept: Convert the line to parametric form, substitute into the plane equation, solve for the parameter.

1. Write line in parametric form:

   \[\frac{x-1}{2} = \frac{y-3}{3} = -z = t\]

   So: \(x = 1 + 2t\), \(y = 3 + 3t\), \(z = -t\)

2. Substitute into plane equation:

   \[3(1 + 2t) - 2(3 + 3t) + 4(-t) = -1\] \[3 + 6t - 6 - 6t - 4t = -1\] \[-3 - 4t = -1\] \[-4t = 2 \quad \Rightarrow \quad t = -\frac{1}{2}\]

3. Find coordinates:

   \[P_0 = \left(1 + 2\left(-\frac{1}{2}\right), 3 + 3\left(-\frac{1}{2}\right), -\left(-\frac{1}{2}\right)\right) = \left(0, \frac{3}{2}, \frac{1}{2}\right)\]

Ответ: \(P_0 = \left(0, \frac{3}{2}, \frac{1}{2}\right)\)

4.95. Плоскость, перпендикулярная прямой (Контрольная, Задание 1b)

Find the equation of a plane perpendicular to line \(L: \frac{x-1}{2} = \frac{y-3}{3} = -z\) through point \(P_0\left(0, \frac{3}{2}, \frac{1}{2}\right)\).

Нажмите, чтобы увидеть решение

Key Concept: The direction vector of the line becomes the normal vector of the plane.

1. Identify direction vector:

   From the line: \(\mathbf{d} = (2, 3, -1)\)

2. Use point-normal form:

   \[2(x - 0) + 3\left(y - \frac{3}{2}\right) - 1\left(z - \frac{1}{2}\right) = 0\]

3. Simplify:

   \[2x + 3y - \frac{9}{2} - z + \frac{1}{2} = 0\] \[2x + 3y - z - 4 = 0\]

Ответ: \(2x + 3y - z - 4 = 0\)

4.96. Прямая, перпендикулярная плоскости (Контрольная, Задание 1c)

Find parametric equations for the line through \(P_0\left(0, \frac{3}{2}, \frac{1}{2}\right)\) perpendicular to plane \(\pi: 3x - 2y + 4z = -1\).

Нажмите, чтобы увидеть решение

Key Concept: The normal vector of the plane becomes the direction vector of the line.

1. Identify normal vector:

   From the plane: \(\mathbf{n} = (3, -2, 4)\)

2. Write parametric equations:

   \[x = 0 + 3t = 3t\] \[y = \frac{3}{2} - 2t\] \[z = \frac{1}{2} + 4t\]

Ответ: \(x = 3t\), \(y = \frac{3}{2} - 2t\), \(z = \frac{1}{2} + 4t\)

4.97. Классификация взаимного расположения прямых (Контрольная, Задание 1d)

Determine the relationship between:

• Line 1: \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{1}\)
• Line 2: \(\frac{x-4}{1} = \frac{y-1}{3} = \frac{z-5}{-2}\)

Нажмите, чтобы увидеть решение

Key Concept: Check if direction vectors are parallel, then check if lines intersect.

1. Identify direction vectors:

   \(\mathbf{d_1} = (2, 3, 1)\) and \(\mathbf{d_2} = (1, 3, -2)\)

2. Check if parallel:

   \(\mathbf{d_1} \neq k\mathbf{d_2}\) for any scalar \(k\), so not parallel.

3. Check if intersecting:

   Set up equations: \(1 + 2t = 4 + s\), \(2 + 3t = 1 + 3s\), \(3 + t = 5 - 2s\)

   From first: \(2t - s = 3\) From second: \(3t - 3s = -1\) From third: \(t + 2s = 2\)

   This system is inconsistent, so lines don’t intersect.

Ответ: The lines are skew (not parallel and not intersecting).

4.98. Плоскости-биссектрисы (Контрольная, Задание 1e)

Find the bisector planes of \(P_1: 2x + 2z + 3 = 0\) and \(P_2: x + y + 2 = 0\).

Нажмите, чтобы увидеть решение

Key Concept: A point lies on a bisector plane if its distances to both planes are equal.

1. Distance formula:

   For a point \((x, y, z)\): \[d_1 = \frac{|2x + 2z + 3|}{\sqrt{2^2 + 0^2 + 2^2}} = \frac{|2x + 2z + 3|}{2\sqrt{2}}\] \[d_2 = \frac{|x + y + 2|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|x + y + 2|}{\sqrt{2}}\]

2. Set equal:

   \[\frac{|2x + 2z + 3|}{2\sqrt{2}} = \frac{|x + y + 2|}{\sqrt{2}}\] \[|2x + 2z + 3| = 2|x + y + 2|\]

3. Case 1: \(2x + 2z + 3 = 2(x + y + 2)\)

   \[2y - 2z + 1 = 0\]

4. Case 2: \(2x + 2z + 3 = -2(x + y + 2)\)

   \[4x + 2y + 2z + 7 = 0\]

Ответ: The bisector planes are \(2y - 2z + 1 = 0\) and \(4x + 2y + 2z + 7 = 0\).

4.99. Смешанное произведение и линейная независимость (Контрольная, Задание 2a)

Given \(\mathbf{a} = (1, 2, 3)\), \(\mathbf{b} = (2, -1, 1)\), \(\mathbf{c} = (3, 1, -2)\), compute the scalar triple product and determine if the vectors are linearly independent.

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Key Concept: The scalar triple product is the determinant of the matrix with vectors as rows (or columns).

1. Calculate scalar triple product:

   \[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{vmatrix}\]

   \[= 1\begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} - 2\begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} + 3\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}\]

   \[= 1(2 - 1) - 2(-4 - 3) + 3(2 + 3)\] \[= 1 + 14 + 15 = 30\]

2. Check linear independence:

   Since the scalar triple product \(\neq 0\), the vectors are linearly independent.

Ответ: Scalar triple product = \(30\); vectors are linearly independent.

4.100. Площадь треугольника через векторное произведение (Контрольная, Задание 2b)

Find the area of triangle with vertices \(A(1, 2, 3)\), \(B(3, 0, 0)\), \(C(5, 2, 9)\).

Нажмите, чтобы увидеть решение

Key Concept: Area = \(\frac{1}{2}||\vec{AB} \times \vec{AC}||\)

1. Find vectors:

   \[\vec{AB} = (3-1, 0-2, 0-3) = (2, -2, -3)\] \[\vec{AC} = (5-1, 2-2, 9-3) = (4, 0, 6)\]

2. Calculate cross product:

   \[\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & -3 \\ 4 & 0 & 6 \end{vmatrix}\]

   \[= \mathbf{i}(-12 - 0) - \mathbf{j}(12 - (-12)) + \mathbf{k}(0 - (-8))\] \[= (-12, -24, 8)\]

3. Find magnitude:

   \[||\vec{AB} \times \vec{AC}|| = \sqrt{144 + 576 + 64} = \sqrt{784} = 28\]

4. Calculate area:

   \[\text{Area} = \frac{1}{2} \times 28 = 14\]

Ответ: \(14\)

4.101. Тождество двойного векторного произведения (Контрольная, Задание 2c)

Verify the identity \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\) for \(\mathbf{a} = (1, 2, 3)\), \(\mathbf{b} = (2, -1, 1)\), \(\mathbf{c} = (3, 1, -2)\).

Нажмите, чтобы увидеть решение

Key Concept: Compute both sides independently and verify equality.

1. Left side - calculate \(\mathbf{b} \times \mathbf{c}\):

   \[\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{vmatrix} = (2 - 1, -(-4 - 3), 2 + 3) = (1, 7, 5)\]

2. Left side - calculate \(\mathbf{a} \times (1, 7, 5)\):

   \[\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & 7 & 5 \end{vmatrix}\] \[= (10 - 21, -(5 - 3), 7 - 2) = (-11, -2, 5)\]

3. Right side - calculate dot products:

   \[\mathbf{a} \cdot \mathbf{c} = 3 + 2 - 6 = -1\] \[\mathbf{a} \cdot \mathbf{b} = 2 - 2 + 3 = 3\]

4. Right side - final calculation:

   \[(\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\] \[= -1(2, -1, 1) - 3(3, 1, -2)\] \[= (-2, 1, -1) + (-9, -3, 6)\] \[= (-11, -2, 5)\]

Ответ: Both sides equal \((-11, -2, 5)\). Identity verified. ✓

4.102. Ранг матрицы (Контрольная, Задание 3a)

Find the rank of matrix \(A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 2 \end{bmatrix}\).

Нажмите, чтобы увидеть решение

Key Concept: Reduce to row echelon form and count non-zero rows.

1. Row reduce:

   \[\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 2 \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 1 & 2 \end{bmatrix}\]

   \[\xrightarrow{R_3 - R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -1 \end{bmatrix}\]

   \[\xrightarrow{\text{swap}} \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\]

2. Count non-zero rows:

   There are 2 non-zero rows.

Ответ: Rank = \(2\)

4.103. Решение системы: метод Гаусса (Контрольная, Задание 3b-i)

Solve the system using Gaussian elimination: \[\begin{cases} x + 2y + 3z = 6 \\ 2x + 4y + 7z = 13 \\ 3x + 7y + 11z = 20 \end{cases}\]

Нажмите, чтобы увидеть решение

Key Concept: Reduce to REF, then use back substitution.

1. Write augmented matrix:

   \[\left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 2 & 4 & 7 & 13 \\ 3 & 7 & 11 & 20 \end{array}\right]\]

2. Row reduce:

   \[\xrightarrow{R_2 - 2R_1} \left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 0 & 0 & 1 & 1 \\ 3 & 7 & 11 & 20 \end{array}\right]\]

   \[\xrightarrow{R_3 - 3R_1} \left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 2 & 2 \end{array}\right]\]

   \[\xrightarrow{\text{swap } R_2, R_3} \left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]\]

3. Back substitution:

   \(z = 1\) \(y + 2(1) = 2 \Rightarrow y = 0\) \(x + 2(0) + 3(1) = 6 \Rightarrow x = 3\)

Ответ: \((3, 0, 1)\)

4.104. Решение системы: Гаусс–Жордан (Контрольная, Задание 3b-ii)

Solve the same system using Gauss-Jordan elimination.

Нажмите, чтобы увидеть решение

Key Concept: Continue from REF to RREF.

1. Start from REF:

   \[\left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]\]

2. Eliminate above pivots:

   \[\xrightarrow{R_1 - 3R_3} \left[\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]\]

   \[\xrightarrow{R_2 - 2R_3} \left[\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right]\]

   \[\xrightarrow{R_1 - 2R_2} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right]\]

Ответ: \((3, 0, 1)\)

4.105. Решение системы: правило Крамера (Контрольная, Задание 3b-iii)

Solve the same system using Cramer’s rule.

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Key Concept: Calculate determinants \(D\), \(D_x\), \(D_y\), \(D_z\).

1. Calculate main determinant:

   \[D = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \\ 3 & 7 & 11 \end{vmatrix} = 1(44-49) - 2(22-21) + 3(14-12) = -5 - 2 + 6 = -1\]

2. Calculate \(D_x\):

   \[D_x = \begin{vmatrix} 6 & 2 & 3 \\ 13 & 4 & 7 \\ 20 & 7 & 11 \end{vmatrix} = 6(44-49) - 2(143-140) + 3(91-80) = -30 - 6 + 33 = -3\]

3. Calculate \(D_y\):

   \[D_y = \begin{vmatrix} 1 & 6 & 3 \\ 2 & 13 & 7 \\ 3 & 20 & 11 \end{vmatrix} = 1(143-140) - 6(22-21) + 3(40-39) = 3 - 6 + 3 = 0\]

4. Calculate \(D_z\):

   \[D_z = \begin{vmatrix} 1 & 2 & 6 \\ 2 & 4 & 13 \\ 3 & 7 & 20 \end{vmatrix} = 1(80-91) - 2(40-39) + 6(14-12) = -11 - 2 + 12 = -1\]

5. Find solution:

   \[x = \frac{D_x}{D} = \frac{-3}{-1} = 3, \quad y = \frac{D_y}{D} = \frac{0}{-1} = 0, \quad z = \frac{D_z}{D} = \frac{-1}{-1} = 1\]

Ответ: \((3, 0, 1)\)

4.106. Angle Between Planes (Контрольная, Задание 4a)

Find the angle between planes with normal vectors \(\mathbf{n_1} = (1, 2, -1)\) and \(\mathbf{n_2} = (3, -1, 2)\).

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Key Concept: Use \(\cos\theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{||\mathbf{n_1}|| ||\mathbf{n_2}||}\).

1. Calculate dot product:

   \[\mathbf{n_1} \cdot \mathbf{n_2} = 3 - 2 - 2 = -1\]

2. Calculate magnitudes:

   \[||\mathbf{n_1}|| = \sqrt{1 + 4 + 1} = \sqrt{6}\] \[||\mathbf{n_2}|| = \sqrt{9 + 1 + 4} = \sqrt{14}\]

3. Find angle:

   \[\cos\theta = \frac{|-1|}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}}\] \[\theta = \cos^{-1}\left(\frac{1}{\sqrt{84}}\right)\]

Ответ: \(\theta = \cos^{-1}\left(\frac{1}{\sqrt{84}}\right) \approx 83.66°\)

4.107. Distance from Point to Plane (Контрольная, Задание 4b)

Find the distance from point \(P(2, -1, 3)\) to plane \(2x - 3y + 6z = 12\).

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Key Concept: Use distance formula \(d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\).

1. Apply formula:

   \[d = \frac{|2(2) - 3(-1) + 6(3) - 12|}{\sqrt{4 + 9 + 36}}\]

   \[= \frac{|4 + 3 + 18 - 12|}{\sqrt{49}}\]

   \[= \frac{13}{7}\]

Ответ: \(\frac{13}{7}\)

4.108. Unique Solution Condition (Контрольная, Задание 4c)

For what values of \(m\) does the system have a unique solution? \[\begin{cases} x + y + z = 1 \\ x + my + z = 2 \\ x + y + mz = 3 \end{cases}\]

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Key Concept: System has unique solution when \(\det(A) \neq 0\).

1. Calculate determinant:

   \[\det(A) = \det\begin{bmatrix} 1 & 1 & 1 \\ 1 & m & 1 \\ 1 & 1 & m \end{bmatrix}\]

   \[= 1(m^2 - 1) - 1(m - 1) + 1(1 - m)\] \[= m^2 - 1 - m + 1 + 1 - m\] \[= m^2 - 2m + 1 = (m - 1)^2\]

2. Find condition:

   For unique solution: \((m - 1)^2 \neq 0 \Rightarrow m \neq 1\)

Ответ: \(m \neq 1\)

4.109. Matrix Inverse (Контрольная, Задание 4d)

Find the inverse of \(B = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 5 & 2 \\ 1 & 3 & 4 \end{bmatrix}\) using Gauss-Jordan method.

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Key Concept: Augment with identity matrix, reduce to \([I|B^{-1}]\).

1. Set up augmented matrix:

   \[\left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 2 & 5 & 2 & 0 & 1 & 0 \\ 1 & 3 & 4 & 0 & 0 & 1 \end{array}\right]\]

2. Row reduce:

   \[\xrightarrow{R_2-2R_1, R_3-R_1} \left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -2 & 1 & 0 \\ 0 & 1 & 3 & -1 & 0 & 1 \end{array}\right]\]

   \[\xrightarrow{R_3-R_2} \left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -2 & 1 & 0 \\ 0 & 0 & 3 & 1 & -1 & 1 \end{array}\right]\]

   \[\xrightarrow{R_3/3} \left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -2 & 1 & 0 \\ 0 & 0 & 1 & \frac{1}{3} & -\frac{1}{3} & \frac{1}{3} \end{array}\right]\]

   Continue to get RREF…

Ответ: \(B^{-1} = \frac{1}{3}\begin{bmatrix} 14 & -5 & -1 \\ -6 & 3 & 0 \\ 1 & -1 & 1 \end{bmatrix}\)

4.110. Determinant of AᵀA and Parallelogram Area (Домашнее задание 6, Задание 1)

Let \(\vec{u} = (u_1, u_2, u_3)\) and \(\vec{v} = (v_1, v_2, v_3)\) be two vectors in \(\mathbb{R}^3\), and consider the parallelogram \(P\) spanned by these vectors. Define the matrix \(A\) by stacking \(\vec{u}\) and \(\vec{v}\) as columns: \(A = [\vec{u} \ \vec{v}] = \begin{bmatrix} u_1 & v_1 \\ u_2 & v_2 \\ u_3 & v_3 \end{bmatrix}\). Prove that the area of the parallelogram \(P\) is given by: Area \(= \sqrt{\det(A^T A)}\).

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1. Area from Cross Product: The area of a parallelogram spanned by two vectors \(\vec{u}\) and \(\vec{v}\) is given by the magnitude of their cross product: Area \(= ||\vec{u} \times \vec{v}||\).
2. Lagrange’s Identity: Squaring the area gives Area\(^2 = ||\vec{u} \times \vec{v}||^2\). By Lagrange’s identity, the squared magnitude of the cross product can be expressed using dot products: \(||\vec{u} \times \vec{v}||^2 = ||\vec{u}||^2 ||\vec{v}||^2 - (\vec{u} \cdot \vec{v})^2\).
3. Compute \(A^T A\): Now let’s compute the determinant of \(A^T A\). First, we find the product \(A^T A\):
   • \(A^T A = \begin{bmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{bmatrix} \begin{bmatrix} u_1 & v_1 \\ u_2 & v_2 \\ u_3 & v_3 \end{bmatrix}\)
   • \(= \begin{bmatrix} u_1^2+u_2^2+u_3^2 & u_1v_1+u_2v_2+u_3v_3 \\ v_1u_1+v_2u_2+v_3u_3 & v_1^2+v_2^2+v_3^2 \end{bmatrix}\)
4. Relate to Dot Products: We can recognize the elements of this matrix as dot products:
   • \(A^T A = \begin{bmatrix} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} \end{bmatrix} = \begin{bmatrix} ||\vec{u}||^2 & \vec{u} \cdot \vec{v} \\ \vec{u} \cdot \vec{v} & ||\vec{v}||^2 \end{bmatrix}\).
5. Calculate the Determinant: Now, we compute the determinant of this \(2 \times 2\) matrix:
   • \(\det(A^T A) = (||\vec{u}||^2)(||\vec{v}||^2) - (\vec{u} \cdot \vec{v})(\vec{u} \cdot \vec{v}) = ||\vec{u}||^2 ||\vec{v}||^2 - (\vec{u} \cdot \vec{v})^2\).
6. Conclusion: The expression for \(\det(A^T A)\) is identical to the expression for Area\(^2\) from Lagrange’s identity in step 2.
   • Therefore, Area\(^2 = \det(A^T A)\).
   • Taking the square root gives Area \(= \sqrt{\det(A^T A)}\).

Ответ: The proof is complete.