W7. Spatial Linear Geometry

Author

Salman Ahmadi-Asl

Published

October 14, 2025

1. Theory

1.1 Equations of Lines in 3D Space

In three-dimensional space, a line can be uniquely determined if we know a point on the line and the direction in which the line extends.

1.1.1 Vector and Parametric Forms

The vector form of a line is an equation that uses a point and a direction vector to represent all points on the line.

  • Let be a known point on the line.
  • Let be a direction vector, which is a non-zero vector parallel to the line.
  • Any arbitrary point on the line can be reached by starting at the origin, moving to point , and then moving along the direction vector by some scalar multiple, .

This relationship is expressed as , which leads to the parametric form of the line: Here, is the parameter, a real number that can vary. As changes, the point traces out the entire line.

1.1.2 Symmetric Equations

If we solve each parametric equation for the parameter (assuming ), we can set them equal to each other to get the symmetric equations of the line: This form is useful for quickly identifying the direction vector and a point on the line. If any component of the direction vector is zero (e.g., ), the line lies on a plane parallel to a coordinate plane, and the symmetric form is adjusted. For instance, if , the equations would be .

1.2 Relationships Between Two Lines

In 3D space, two lines can relate in four distinct ways:

  1. Parallel Lines: The lines have the same direction (their direction vectors are scalar multiples of each other) but never intersect.
  2. Intersecting Lines: The lines are not parallel and cross at exactly one point.
  3. Skew Lines: The lines are not parallel and do not intersect. They exist in different, non-intersecting planes. Think of two airplanes flying at different altitudes on different paths that cross on a map but never touch.
  4. Coincident Lines: The lines are identical; they are the same line. Their direction vectors are scalar multiples, and they share at least one point.
1.3 Distance from a Point to a Line

The shortest distance from a point to a line is the length of the perpendicular segment connecting them.

  • Let be the point not on the line.
  • Let the line be defined by a point and a direction vector .
  • The distance can be found using the cross product. The magnitude of the cross product of the vector and the direction vector gives the area of a parallelogram. Dividing this area by the length of the base () gives the height, which is the desired distance.

The formula is:

1.4 Equations of Planes in 3D Space

A plane is a flat, two-dimensional surface that extends infinitely in 3D space. It can be defined by a point on the plane and a vector perpendicular to it.

1.4.1 Point-Normal Form

A plane is uniquely determined by a point lying in the plane and a normal vector that is orthogonal (perpendicular) to every vector in the plane.

For any other point in the plane, the vector must be orthogonal to the normal vector . Since the dot product of two orthogonal vectors is zero, we have . This gives the point-normal form:

1.4.2 Standard (General) Form

By expanding the point-normal form and rearranging terms, we get the standard form or general equation of a plane: where the coefficients are the components of the normal vector , and .

1.5 Distance from a Point to a Plane

The distance from a point to the plane (or ) is the length of the perpendicular line segment from the point to the plane. The formula is: The numerator is the absolute value of the plane’s equation evaluated at the point’s coordinates. The denominator is the magnitude of the normal vector.

1.6 Angle Between Two Planes

The angle between two intersecting planes is defined as the acute angle between their normal vectors.

  • Let the planes be and , with normal vectors and .
  • The angle between the planes can be found using the dot product formula for the angle between two vectors.

1.7 Linear Systems and Geometric Interpretation

Systems of linear equations have a direct geometric interpretation in terms of lines and planes.

1.7.1 Lines as Solutions to Systems (2D)

A system of two linear equations in two variables, such as: represents two lines in a 2D plane. The solution to the system is the set of points where the lines intersect.

  • Unique Solution: The lines intersect at a single point.
  • No Solution: The lines are parallel and distinct.
  • Infinitely Many Solutions: The lines are coincident (the same line).
1.7.2 Planes as Solutions to Systems (3D)

A system of three linear equations in three variables represents three planes in 3D space. The solution set is the geometric intersection of these three planes.

  • Unique Solution: The three planes intersect at a single point.
  • Infinite Solutions: The three planes intersect along a common line.
  • No Solution: The planes do not share any common points. This can happen if at least two planes are parallel and distinct, or if the planes form a triangular prism.

2. Definitions

  • Direction Vector: A non-zero vector that is parallel to a line in space. It defines the line’s orientation.
  • Normal Vector: A vector that is orthogonal (perpendicular) to a plane. It defines the plane’s orientation.
  • Parametric Equations of a Line: A set of equations () that define the coordinates of points on a line in terms of a single parameter .
  • Symmetric Equations of a Line: An equation of the form that relates the coordinates of points on a line.
  • Skew Lines: Two lines in 3D space that are not parallel and do not intersect.
  • Point-Normal Form: The equation of a plane, , derived from a point on the plane and its normal vector.
  • Standard Form of a Plane: The equation of a plane written as .

3. Formulas

  • Parametric Equations of a Line:
  • Symmetric Equations of a Line:
  • Point-Normal Form of a Plane:
  • Standard Form of a Plane:
  • Normal Vector from Cross Product:
  • Dot Product:
  • Cross Product:
  • Angle Between Two Vectors:
  • Angle Between Two Planes:
  • Distance from a Point to a Line:
  • Distance from a Point to a Plane:
  • Distance Between Skew Lines:
  • Scalar Triple Product (Volume of Parallelepiped):
  • Area of a Triangle in 3D:
  • Area of Parallelogram from Determinant:
  • Vector Projection:
  • Vector Reflection:
  • Vector Triple Product Identity:
  • Linear Independence Condition (Determinant):
  • Unit Vector (Normalization):
  • Intercept Form of a Line (2D):
  • Midpoint Formula (2D):
  • Midpoint of a Segment (3D):
  • Slope of a Line (2D):
  • Slope from Angle:
  • Point-Slope Form (2D):
  • Perpendicular Slopes (2D):
  • Centroid of a Triangle:
  • Section Formula:
  • Matrix Inverse by Adjugate Method:
  • Singular Matrix Condition:
  • Determinant of a Product:
  • Cramer’s Rule:
  • Condition for Perpendicularity:
  • Angle Between a Line and a Plane:
  • Condition for Unique Solution or Nonsingular Matrix:
  • Condition for Nontrivial Solutions in a Homogeneous System:
  • Magnitude of Cross Product (Geometric):
  • Dot Product (Geometric):
  • Magnitude of Vector Difference Squared:
  • Distance from a Point to a Line in 3D:

4. Practice

4.1. Find the Equation of a Straight Line (Lab 6, Task 1)

Find the equation of the straight line, the portion of which between the axes is bisected at the point .

Click to see the solution
  1. Use the Intercept Form: Let the equation of the line be , where the line intersects the x-axis at and the y-axis at .
  2. Use the Midpoint Formula: The midpoint of the segment between the intercepts is given by .
  3. Find the Intercepts: We are given that the midpoint is .
    • .
    • .
  4. Write the Equation: Substitute the values of and into the intercept form.
    • .
  5. Simplify to General Form: Multiply the entire equation by 20 to clear the denominators.
    • .
    • .
Answer: The equation of the line is .
4.2. Find the Equation of a Straight Line (Lab 6, Task 2)

Find the equation of the straight line passing through the intersection of the lines and and making an angle of with the positive direction of the x-axis.

Click to see the solution
  1. Find the Point of Intersection: Solve the system of two linear equations.
    • From , we get .
    • Substitute this into the second equation: .
    • .
    • Substitute back to find y: .
    • The point of intersection is .
  2. Find the Slope: The slope of a line is the tangent of the angle it makes with the positive x-axis.
    • .
  3. Use the Point-Slope Form: The equation of a line with slope passing through is .
    • .
  4. Simplify to General Form:
    • .
    • .
Answer: The equation of the line is .
4.3. Find the Equation of an Altitude of a Triangle (Lab 6, Task 3)

, , and are the vertices of a triangle. Find the line equation which goes from A and is perpendicular to BC.

Click to see the solution
  1. Find the Slope of BC:
    • .
  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.
    • .
  3. Use the Point-Slope Form: The line passes through point with a slope of .
    • .
  4. Simplify to General Form:
    • .
    • .
    • .
Answer: The equation of the altitude from A is .
4.4. Find the Centroid of a Triangle (Lab 6, Task 4)

Find the centroid of the triangle formed by the lines given by the equations and .

Click to see the solution
  1. Find the Equations of the Three Lines:
    • The equation represents a pair of straight lines passing through the origin. We can factor it:
    • .
    • The three lines forming the triangle are:
  2. Find the Vertices of the Triangle:
    • Vertex 1 (Intersection of and ): Both lines pass through the origin, so their intersection is .
    • Vertex 2 (Intersection of and ): Substitute into : . Then . The vertex is .
    • Vertex 3 (Intersection of and ): Substitute into : . Then . The vertex is .
  3. Calculate the Centroid: The centroid is the average of the coordinates of the vertices .
    • .
    • .
Answer: The centroid of the triangle is .
4.5. Find the Equation of a Parallel Plane (Lab 6, Task 5)

Find the equation of the plane passing through the point and parallel to the plane .

Click to see the solution
  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.
    • .
  2. Write the General Equation of the New Plane: The equation of any plane parallel to the given one is of the form .
  3. Find the Constant : The plane must pass through the point . Substitute these coordinates into the equation to solve for .
    • .
    • .
    • .
  4. Write the Final Equation:
    • .
Answer: The equation of the plane is .
4.6. Find the Equation of a Plane Perpendicular to Two Other Planes (Lab 6, Task 6)

Find the equation of the plane passing through the point and perpendicular to the planes and .

Click to see the solution
  1. Find the Normal Vectors of the Given Planes:
    • .
    • .
  2. Find the Normal Vector of the New Plane: The normal vector of the desired plane must be perpendicular to both and . We can find this by calculating their cross product.
    • .
    • So, .
  3. Use the Point-Normal Form: Use the calculated normal vector and the given point .
    • .
  4. Simplify to General Form:
    • .
    • .
    • Multiplying by -1 gives .
Answer: The equation of the plane is .
4.7. Section Formula for a Line and Plane (Lab 6, Task 7)

Find the ratio in which the line joining the points and is divided by the plane .

Click to see the solution
  1. Use the Section Formula: Let the point of intersection R divide the line segment PQ in the ratio . The coordinates of R are given by .
  2. Substitute into the Plane Equation: Since R lies on the plane, its coordinates must satisfy the plane’s equation.
    • .
  3. Solve for : Multiply the entire equation by to clear the denominator.
    • .
    • .
    • .
    • .
  4. Interpret the Ratio: The ratio is , which is or . The negative sign indicates that the plane divides the line segment PQ externally.
Answer: The plane divides the line segment in the ratio -17:27 (externally).
4.8. Perpendicular Distance from a Point to a Line (Lab 6, Task 8)

Find the perpendicular distance from the point to the line .

Click to see the solution
  1. Identify Information from the Line:
    • A point on the line is .
    • The direction vector of the line is .
  2. Find the Vector :
    • .
  3. Use the Distance Formula: The distance is .
  4. Calculate the Cross Product:
    • .
  5. Calculate the Magnitudes:
    • .
    • .
  6. Calculate the Distance:
    • .
Answer: The distance is .
4.9. Distance from Points to a Plane (Lab 6, Task 9)

Find the distance of the points and from the plane and prove that these points lie on the opposite sides of the plane.

Click to see the solution
  1. Rewrite the Plane Equation: Let .
  2. Calculate the Distances: Use the distance formula .
    • The denominator is .
    • For point : .
    • For point : .
  3. Check if Points are on Opposite Sides: Substitute the coordinates of each point into the expression . If the signs of the results are different, the points are on opposite sides.
    • (positive).
    • (negative).
  4. Conclusion: Since one result is positive and the other is negative, the points lie on opposite sides of the plane.
Answer: The distance for is 3. The distance for 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. Angle Between Two Planes (Lab 6, Task 10)

Given two planes and where and are non-zero vectors, find the angle between these two planes.

Click to see the solution
  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:
  3. Dot Product Formula: The cosine of the angle between two vectors is given by the dot product formula: .
  4. Acute Angle: To ensure the angle is acute (between and ), we take the absolute value of the dot product.
    • .
  5. Final Formula: The angle is the arccosine of this value.
    • .
Answer: The angle between the two planes is given by the formula .
4.11. Compute Matrix Inverse (Lab 6, Task 11)

Given , compute the inverse of A using:

  1. Adjugate method
  2. Gauss-Jordan method
Click to see the solution

(a) Adjugate Method

  1. Calculate the Determinant:
    • .
  2. Find the Matrix of Cofactors:
    • .
  3. Find the Adjugate Matrix: The adjugate is the transpose of the cofactor matrix.
    • .
  4. Find the Inverse: .
    • .

(b) Gauss-Jordan Method

  1. Set up the Augmented Matrix :
  2. Perform Row Operations to get RREF:
    • , :
    • :
    • :
Answer: Both methods yield the inverse .
4.12. Compute Matrix Inverse (Lab 6, Task 12)

Given , compute the inverse of B using:

  1. Adjugate method
  2. Gauss-Jordan method
Click to see the solution

(a) Adjugate Method

  1. Calculate the Determinant:
    • .
  2. Find the Matrix of Cofactors:
    • .
  3. Find the Adjugate Matrix:
    • .
  4. Find the Inverse: .

(b) Gauss-Jordan Method

  1. Set up and reduce:
Answer: Both methods yield the inverse .
4.13. Classify Echelon Forms (Lab 6, Task 13)

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

Click to see the solution
  1. (a) : 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) : Neither. The leading entry of row 3 (2) is not to the right of the leading entry of row 2 (1).
  3. (c) : 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) : 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) : Neither. The row of all zeros is not at the bottom of the matrix.
  6. (f) : Both REF and RREF. The pivots (the two 1s) are the only non-zero entries in their columns.
4.14. Solve a Linear System by Multiple Methods (Lab 6, Task 14)

Given the system , solve using:

  1. Gaussian elimination
  2. Gauss-Jordan elimination
  3. Cramer’s rule
Click to see the solution

(a) Gaussian Elimination

  1. Row reduce the augmented matrix to REF:
    • .
  2. Back Substitution:
    • From R3: .
    • From R2: .
    • From R1: .

(b) Gauss-Jordan Elimination

  1. Continue row reduction from REF to RREF:
    • .
  2. Read the Solution: .

(c) Cramer’s Rule

  1. Calculate Determinants:
    • .
    • .
    • .
    • .
  2. Find the Solution:
    • .
    • .
    • .
Answer: The unique solution is .
4.15. Solve a Homogeneous Linear System (Lab 6, Task 15)

Given the system , solve using:

  1. Gaussian elimination
  2. Gauss-Jordan elimination
Click to see the solution

(a) Gaussian Elimination

  1. Row reduce the coefficient matrix to REF:
    • .
  2. Back Substitution: The free variable is . Let .
    • From R4: .
    • From R3: .
    • From R2: .
    • From R1: .

(b) Gauss-Jordan Elimination

  1. Continue row reduction from REF to RREF:
    • .
  2. Read the Solution:
    • .
    • .
    • .
    • .
    • Let .
Answer: The general solution is for any scalar .
4.16. Find the Equation of a Plane (Lecture 6, Example 1)

Find the equation of the plane passing through the point with a normal vector of .

Click to see the solution
  1. Use the Point-Normal Form: The equation of a plane is given by , where is the normal vector and is a point on the plane.
  2. Substitute the given values: We have and .
    • .
  3. Simplify to the General Form: Expand and simplify the equation.
    • .
    • .
    • .
Answer: The equation of the plane is .
4.17. Find the Normal Vector of a Plane (Lecture 6, Example 2)

Given the plane , find its normal vector.

Click to see the solution
  1. Recall the General Form: The general equation of a plane is .
  2. Identify the Coefficients: The coefficients of and in the general form represent the components of a vector normal (perpendicular) to the plane.
  3. Extract the Normal Vector: For the equation , the coefficients are .
    • The normal vector is .
Answer: A normal vector to the plane is .
4.18. Find the Line of Intersection of Two Planes (Lecture 6, Example 3)

Find the line of intersection of the planes:

Click to see the solution
  1. Find the Direction Vector: The line of intersection lies in both planes, so its direction vector must be perpendicular to both normal vectors, and . We can find by taking the cross product.
    • . We can use a simpler parallel vector, like .
  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 .
  3. Solve the 2x2 System:
    • From the second equation, .
    • Substitute into the first: .
    • Now find : .
    • So, a point on the line is .
  4. Write the Equation of the Line: Using the point and the direction vector , the symmetric equation is:
    • .
Answer: The line of intersection is given by the symmetric equations .
4.19. Vector Operations (Lecture 6, Example 4)

Given vectors and , find:

  1. The dot product
  2. The cross product
  3. The angle between and
Click to see the solution

First, write the vectors in component form: and .

(a) The dot product

  1. Use the dot product formula: .
  2. Calculate:
    • .
    • Answer (a): .

(b) The cross product

  1. Set up the determinant:
  2. Calculate:
    • .
    • Answer (b): .

(c) The angle between and

  1. Calculate the magnitudes:
    • .
    • .
  2. Use the angle formula: .
  3. Calculate:
    • .
    • .
    • Answer (c): .
4.20. Find the Equation of a Plane Through Three Points (Lecture 6, Example 5)

Find the equation of the plane passing through the points , , and .

Click to see the solution
  1. Find Two Vectors in the Plane: Let the points be .
    • .
    • .
  2. Find the Normal Vector: The normal vector is the cross product of these two vectors.
    • .
    • So, .
  3. Use the Point-Normal Form: Using point and .
    • .
  4. Simplify:
    • .
    • .
Answer: The equation of the plane is .
4.21. Determine if Lines are Parallel (Lecture 6, Example 6)

Consider the two lines: Explain why they are parallel.

Click to see the solution
  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 is .
    • Direction vector of is .
  2. Check for Proportionality: Two lines are parallel if their direction vectors are scalar multiples of each other, i.e., for some scalar .
  3. Compare the Vectors: Let’s see if we can find a such that .
    • From the first component: .
    • Check with the other components:
      • Is ? Yes, .
      • Is ? Yes, .
  4. Conclusion: Since , the direction vectors are parallel, and therefore the lines are parallel.
Answer: The lines are parallel because their direction vectors, and , are scalar multiples of each other.
4.22. Scalar Triple Product (Lecture 6, Example 7)

Find the scalar triple product of vectors , , and . What does the result tell you about these vectors?

Click to see the solution
  1. Set up the determinant: The scalar triple product is the determinant of the matrix formed by the three vectors.
  2. Calculate the determinant:
    • .
  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).
Answer: The scalar triple product is 30. This means the vectors are not coplanar.
4.23. Determine if Lines are Parallel, Intersecting, or Skew (Lecture 6, Example 8)

Determine whether the lines and are parallel, intersecting, or skew.

Click to see the solution
  1. Check for Parallelism: Compare the direction vectors.
    • .
    • .
    • 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.
    • : .
    • : .
  3. Set up and solve the system:
    • .
    • .
    • .
    • Substitute into the second equation: .
    • Find : .
    • Check these values in the third equation: . This is a contradiction.
  4. Conclusion: The lines are not parallel and they do not intersect.
Answer: The lines are skew.
4.24. Determine if Lines are Parallel (Lecture 6, Example 9)

Consider the following lines: Explain why they are not parallel.

Click to see the solution
  1. Identify the Direction Vectors:
    • Direction vector of is .
    • Direction vector of is .
  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 such that .
  3. Compare the Components:
    • From the x-component: .
    • Check this with the y-component: Is ? No, .
  4. Conclusion: Since the components are not proportional, the direction vectors are not scalar multiples of each other.
Answer: The lines are not parallel because their direction vectors, and , are not scalar multiples of each other.
4.25. Vector Projection (Lecture 6, Example 10)

Find the projection of vector onto vector .

Click to see the solution
  1. Use the projection formula: .
  2. Calculate the dot product:
    • .
  3. Calculate the squared magnitude of :
    • .
  4. Calculate the projection vector:
    • .
Answer: The projection of onto is .
4.26. Distance from a Point to a Plane (Lecture 6, Example 11)

Find the distance from the point to the plane .

Click to see the solution
  1. State the Distance Formula: The distance from to is .
  2. Identify the values:
    • Point: .
    • Plane: , so .
  3. Substitute and calculate:
    • .
Answer: The distance is .
4.27. Determine if Lines are Skew (Lecture 6, Example 12)

Consider two lines: Explain why they are skew.

Click to see the solution
  1. Check for Parallelism:
    • Direction vector of : .
    • Direction vector of : .
    • 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.
    • : , , .
    • : , , .
  3. Set the coordinates equal to find a potential intersection:
  4. Solve the system of equations:
    • From the second equation, .
    • Substitute this into the first equation: .
    • Using , we find .
    • Now check if these values of and satisfy the third equation: Is ?
    • . 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.
Answer: The lines are skew because their direction vectors are not parallel and there is no point of intersection.
4.28. Vector Reflection (Lecture 6, Example 13)

Find the reflection of vector about the vector .

Click to see the solution
  1. State the Formula: The reflection of a vector across the line defined by vector is given by the formula .
  2. Calculate the projection of onto :
    • .
    • .
    • .
  3. Calculate the reflection vector:
    • .
Answer: The reflection is .
4.29. Acute Angle Between Planes (Lecture 6, Example 14)

Find the acute angle between the planes and .

Click to see the solution
  1. Find the Normal Vectors:
    • .
    • .
  2. Use the Dot Product Formula: .
  3. Calculate the components:
    • .
    • .
    • .
  4. Calculate the angle:
    • .
    • .
Answer: The acute angle is .
4.30. Angle Between Two Planes (Lecture 6, Example 15)

Find the angle between the planes and .

Click to see the solution
  1. Identify the Normal Vectors: The angle between two planes is the angle between their normal vectors.
    • Normal vector of the first plane: .
    • Normal vector of the second plane: .
  2. Use the Dot Product Formula: The cosine of the angle between two vectors is given by .
  3. Calculate the Dot Product:
    • .
  4. Calculate the Magnitudes:
    • .
    • .
  5. Calculate the Cosine of the Angle:
    • .
    • .
Answer: The angle between the planes is , which is approximately .
4.31. Triangle Area and Parallelepiped Volume (Lecture 6, Example 16)

Given points , , and , find:

  1. The area of triangle
  2. The volume of the parallelepiped formed by vectors , , and where
Click to see the solution

First, find the vectors originating from point A:

(a) Area of triangle

  1. Calculate the cross product: Area .
    • .
  2. Find the magnitude:
    • .
  3. Calculate the area:
    • Area .
    • Answer (a): .

(b) Volume of the parallelepiped

  1. Calculate the scalar triple product: Volume .
  2. Calculate:
    • .
    • Answer (b): .
4.32. Symmetric Equations of a Line (Lecture 6, Example 17)

Find the symmetric equations of the line through perpendicular to the plane .

Click to see the solution
  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: .
    • Direction vector of the line: .
  2. Use the Symmetric Form: The symmetric equations for a line through with direction are .
  3. Substitute the values:
    • Point: .
    • Direction: .
    • .
Answer: The symmetric equations are .
4.33. Linear Independence of Vectors (Lecture 6, Example 18)

Determine if the vectors , , and are linearly independent.

Click to see the solution
  1. Use the determinant: Three vectors in are linearly independent if and only if the determinant of the matrix formed by them is non-zero.
  2. Set up the determinant:
  3. Calculate the determinant:
    • .
  4. Conclusion: Since the determinant is 1 (non-zero), the vectors are linearly independent.
Answer: Yes, the vectors are linearly independent.
4.34. Intersection of a Line and a Plane (Lecture 6, Example 19)

Determine the point of intersection of the line with the plane .

Click to see the solution
  1. Write the Line in Parametric Form:
    • Let .
  2. Substitute into the Plane Equation:
    • .
  3. Solve for :
    • .
    • .
  4. Find the Intersection Point: Substitute back into the parametric equations.
    • .
    • .
    • .
Answer: The point of intersection is .
4.35. Distance from a Point to a Plane (Lecture 6, Example 20)

Find the distance from the point to the plane .

Click to see the solution
  1. State the Distance Formula: The distance from a point to a plane is given by:
    • .
  2. Identify the Values:
    • Point: .
    • Plane coefficients: .
  3. Substitute into the Formula:
    • .
  4. Calculate the Result:
    • .
Answer: The distance is .
4.36. Find a Perpendicular Unit Vector (Lecture 6, Example 21)

Find a unit vector perpendicular to both and .

Click to see the solution
  1. Find the cross product: A vector perpendicular to both and is their cross product.
    • .
  2. Find the magnitude of the cross product:
    • .
  3. Normalize the vector: Divide the vector by its magnitude to get a unit vector.
    • .
Answer: The unit vector is .
4.37. Plane Containing Parallel Lines (Lecture 6, Example 22)

Find the equation of the plane containing the parallel lines: and .

Click to see the solution
  1. Find a Direction Vector: The common direction vector is .
  2. Find a Vector Connecting the Lines: Pick one point from each line.
    • from the first line.
    • from the second line.
    • The vector connecting them is .
  3. Find the Normal Vector to the Plane: The normal vector is the cross product of the two vectors in the plane.
    • .
  4. Write the Plane Equation: Use the point-normal form with point and normal .
    • .
    • .
    • .
Answer: The equation of the plane is .
4.38. Distance from a Point to a Line (Lecture 6, Example 23)

Compute the distance of the point from a line with parametric equations:

Click to see the solution
  1. State the Distance Formula: The distance from a point to a line is given by , where is any point on the line and is the direction vector of the line.
  2. Identify a Point and Direction Vector from the Line:
    • By setting , we get a point on the line: .
    • The direction vector is given by the coefficients of : .
  3. Find the Vector :
    • .
  4. Calculate the Cross Product :
    • .
  5. Calculate the Magnitudes:
    • .
    • .
  6. Calculate the Distance:
    • .
Answer: The distance from the point to the line is .
4.39. Verify Vector Identity (Lecture 6, Example 24)

Verify the vector identity for vectors , , .

Click to see the solution
  1. Calculate the Left-Hand Side (LHS):
    • First, find .
    • Then, calculate .
  2. Calculate the Right-Hand Side (RHS):
    • First, calculate the dot products:
      • .
      • .
    • Then, calculate the full expression: .
    • .
  3. Compare: The LHS is equal to the RHS .
Answer: The identity is verified.
4.40. Distance Between Skew Lines (Lecture 6, Example 25)

Find the distance between the skew lines: and .

Click to see the solution
  1. Identify Points and Direction Vectors:
    • : , .
    • : , .
  2. Find the Common Normal Vector: .
    • .
  3. Find the Vector Connecting the Lines: .
  4. Use the Distance Formula: .
  5. Calculate:
    • .
    • .
    • .
Answer: The distance is .
4.41. Distance from a Point to a Line (Lecture 6, Example 26)

Find the distance from the point to the line through points and .

Click to see the solution
  1. Find the direction vector of the line: .
  2. Find the vector from a point on the line to P: .
  3. Use the distance formula: .
  4. Calculate the cross product:
    • .
  5. Calculate the magnitudes:
    • .
    • .
  6. Calculate the distance: .
Answer: The distance is .
4.42. Reflection of a Point in a Plane (Lecture 6, Example 27)

Find the reflection of the point in the plane .

Click to see the solution
  1. Find the Line Through the Point Perpendicular to the Plane:
    • The direction vector is the plane’s normal: .
    • The line is .
  2. Find the Intersection of the Line and Plane: Substitute the line’s equations into the plane’s equation.
    • .
    • .
    • The intersection point is: , , . So .
  3. Find the Reflection: The intersection is the midpoint of the original point and its reflection .
    • Let . Then , , .
    • .
    • .
    • .
Answer: The reflection is .
4.43. Determine if Four Points are Coplanar (Lecture 6, Example 28)

Determine whether the four points are coplanar.

Click to see the solution
  1. Form Three Vectors: Create three vectors from the points, using as the origin.
    • .
    • .
    • .
  2. Calculate the Scalar Triple Product: The points are coplanar if the volume of the parallelepiped formed by these vectors is zero.
    • Volume
    • .
  3. Conclusion: Since the volume is not zero, the vectors are not coplanar.
Answer: The points are not coplanar.
4.44. Determinant of a Matrix (Lecture 6, Example 29)

Find the determinant of the matrix: .

Click to see the solution
  1. Use the cofactor expansion method along the first row:
  2. Calculate the determinants of the matrices:
    • .
Answer: The determinant is 1.
4.45. Find the Rank of a Matrix (Lecture 6, Example 30)

Find the rank of the matrix: .

Click to see the solution
  1. Use Gaussian Elimination: Reduce the matrix to row echelon form.
    • .
  2. Swap Rows: Swap and to get closer to echelon form.
    • .
  3. Count Non-Zero Rows: The matrix is now in row echelon form. There are two non-zero rows.
Answer: The rank of the matrix is 2.
4.46. Find Matrix Inverse using Adjugate Method (Lecture 6, Example 31)

Find the inverse of the following matrices using the adjugate method: ,

Click to see the solution

For Matrix A:

  1. Calculate the determinant: .
  2. Find the matrix of cofactors: .
  3. Find the adjugate (transpose of cofactors): .
  4. Calculate the inverse: .

For Matrix B:

  1. Calculate the determinant: .
  2. Find the matrix of cofactors: .
  3. Find the adjugate: .
  4. Calculate the inverse: .
Answer:
4.47. Solve a System using Gaussian Elimination (Lecture 6, Example 32)

Use Gaussian elimination to solve the system:

Click to see the solution
  1. Write the Augmented Matrix:
  2. Perform Row Operations:
    • and
  3. Swap Rows: Swap and .
  4. Back Substitution:
    • .
    • .
    • .
Answer: The solution is .
4.48. Calculate Determinant using Properties (Lecture 6, Example 33)

Calculate the determinant using properties of determinants: .

Click to see the solution
  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.
    • and .
  2. Analyze the Result: Notice that the third row is exactly 2 times the second row ().
  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.
Answer: The determinant is 0.
4.49. Find Matrix Inverse using Gauss-Jordan (Lecture 6, Example 34)

Find the inverse of the matrix using Gauss-Jordan elimination: .

Click to see the solution
  1. Set up the Augmented Matrix :
  2. Create Zeros Below the First Pivot:
    • ,
  3. Create Zero Below the Second Pivot:
  4. Create Zeros Above the Pivots:
    • :
    • :
    • :
Answer: .
4.50. Find Values for a Singular Matrix (Lecture 6, Example 35)

For what value(s) of is the matrix singular? .

Click to see the solution
  1. Condition for Singularity: A matrix is singular if its determinant is zero.
  2. Calculate the determinant in terms of :
    • .
  3. Set the determinant to zero and solve:
    • .
  4. Find the values of :
    • or .
Answer: The matrix is singular when or .
4.51. Column Space and Null Space (Lecture 6, Example 36)

Determine the column space and null space of: .

Click to see the solution
  1. Find the Row Echelon Form:
    • .
  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: .
  3. Determine the Null Space: Solve using the RREF.
    • . is the pivot variable, are free.
    • .
    • The solution vector is .
    • Basis for Null Space: .
Answer: The column space is the line spanned by . The null space is the plane spanned by and .
4.52. Verify Matrix Inverse (Lecture 6, Example 37)

Verify that for matrix using your result from question 12.

Click to see the solution
  1. Recall the Inverse: From question 12, .
  2. Perform the multiplication :
  3. Simplify the result:
    • .
Answer: The product is the identity matrix , so the inverse is correct.
4.53. Analyze a System with a Parameter (Lecture 6, Example 38)

For what values of does the system have: (a) A unique solution? (b) Infinitely many solutions? (c) No solution?

Click to see the solution
  1. Row Reduce the Augmented Matrix:
    • :
  2. Analyze the Last Row:
    • The last row represents the equation .
    • .
  3. Determine the Conditions:
    • (c) No solution: The system is inconsistent if the last row is . This happens if . So, no solution if and .
    • (b) Infinitely many solutions: The system has infinite solutions if the last row is . This happens if . So, infinite solutions if or .
    • (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.
Answer: (a) Never, (b) or , (c) and .
4.54. Solve System using Matrix Inversion (Lecture 6, Example 39)

Solve the system using matrix inversion:

Click to see the solution
  1. Write in Matrix Form: , where , , and .
  2. Use the Inverse: The solution is . The matrix A is the same as in question 12.
    • .
  3. Calculate the solution:
  4. Simplify:
    • .
Answer: The solution is .
4.55. Linear Independence and Basis (Lecture 6, Example 40)

Determine whether the following vectors are linearly independent and find a basis for their span: .

Click to see the solution
  1. Check for Linear Independence: We can check if the determinant of the matrix formed by these vectors is non-zero.
    • .
  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 , they already form a basis for their span. Their span is the entire space .
Answer: The vectors are linearly independent. A basis for their span is the set of vectors themselves: .
4.56. Find the Determinant of a Product (Lecture 6, Example 41)

Find the determinant of the product: .

Click to see the solution
  1. Use the Property : We can find the determinant of each matrix and then multiply the results.
  2. Calculate Determinant of First Matrix (A):
    • .
  3. Calculate Determinant of Second Matrix (B):
    • .
  4. Multiply the Determinants:
    • .
Answer: The determinant of the product is 20.
4.57. Cramer’s Rule (Lecture 6, Example 42)

Use Cramer’s rule to solve:

Click to see the solution
  1. Calculate the Main Determinant (D):
    • .
  2. Calculate :
    • .
  3. Calculate :
    • .
  4. Calculate :
    • .
  5. Find the Solution:
    • .
    • .
    • .
Answer: .
4.58. Prove (Lecture 6, Example 43)

Prove that for any matrix , .

Click to see the solution
  1. Define a general 3x3 matrix A:
  2. Calculate :
    • .
  3. Find the transpose :
  4. Calculate :
    • .
    • .
  5. Compare the expanded forms:
    • .
    • The terms are identical.
Answer: By expanding the determinants of a general matrix and its transpose, we see that the resulting expressions are identical, thus proving .
4.59. Rank and Invertibility (Lecture 6, Example 44)

Prove that if is an matrix with rank , then is invertible.

Click to see the solution
  1. Definition of Rank: An matrix has rank if and only if its columns (and rows) are linearly independent.
  2. Implication of Linear Independence: If the columns of are linearly independent, the only solution to the homogeneous equation is the trivial solution .
  3. Connection to Row Echelon Form: If the only solution to is the trivial one, it means that when we reduce the augmented matrix to row echelon form, there are no free variables. For an matrix, this means there must be a pivot in every column.
  4. Reduced Row Echelon Form: An matrix with a pivot in every column can be row-reduced to the identity matrix, .
  5. Definition of Invertibility: A square matrix is invertible if and only if its reduced row echelon form is the identity matrix. Since we have shown that implies that the RREF of A is , it follows that A is invertible.
Answer: A rank of for an 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. Find the Inverse of an Elementary Matrix (Lecture 6, Example 45)

Find the inverse of the elementary matrix and verify that .

Click to see the solution
  1. Interpret the Matrix Operation: The matrix 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:
    • .
  4. Verify the Inverse: Multiply by .
    • .
Answer: . The verification shows that .
4.61. Distance Between Skew Lines (Tutorial 6, Example 1)

Consider two skew lines and . What is the smallest distance between these two lines?

Click to see the solution
  1. Identify Points and Direction Vectors:
    • For : A point on the line is and the direction vector is .
    • For : A point on the line is and the direction vector is .
  2. Find a Common Normal Vector: The vector perpendicular to both lines is the cross product of their direction vectors.
    • .
  3. Find the Vector Connecting the Lines: Create a vector from a point on one line to a point on the other.
    • .
  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.
    • .
  5. Calculate:
    • .
    • .
    • .
Answer: The smallest distance between the two lines is .
4.62. Perpendicular Bisector Plane (Tutorial 6, Example 2)

Find the equation of the plane that is the perpendicular bisector of the segment joining the points and .

Click to see the solution
  1. Find the Normal Vector: The normal vector to the plane is the vector connecting the two points.
    • .
  2. Find the Midpoint: The plane must pass through the midpoint of the segment.
    • .
  3. Use the Point-Normal Form: Use the normal vector and the midpoint .
    • .
  4. Simplify the Equation:
    • .
    • Multiply by 2 to clear the fraction: .
    • .
Answer: The equation of the plane is .
4.63. Line and Plane Interactions (Tutorial 6, Example 3)

Consider the line and the plane .

  1. Find the point of intersection.
  2. Find an equation for the plane perpendicular to that passes through .
  3. Find parametric equations for the line that passes through and is perpendicular to .
Click to see the solution
  1. Find the Point of Intersection :
    • First, write the line in parametric form: .
    • , , .
    • Substitute into the plane equation: .
    • .
    • Substitute back to find the point: , , .
    • .
  2. Find the Perpendicular Plane:
    • The direction vector of the line , , is the normal vector for the new plane.
    • Using the point-normal form with : .
    • .
  3. Find the Perpendicular Line:
    • The normal vector of the plane , , is the direction vector for the new line.
    • The line passes through .
    • The parametric equations are: , , .

Answer:

  1. The point of intersection is .
  2. The equation of the perpendicular plane is .
  3. The parametric equations of the perpendicular line are .
4.64. Line of Intersection of Two Planes (Tutorial 6, Example 4)

Find the line of intersection of the planes given by and .

Click to see the solution
  1. Find the Direction Vector: The direction vector of the line is the cross product of the normal vectors of the planes.
    • and .
    • .
    • We can use the simpler parallel vector .
  2. Find a Point on the Line: Set one variable to 0 and solve the system. Let .
    • .
    • .
    • A point on the line is .
  3. Write the Equation of the Line: Using the point and the direction vector .
    • Parametric: .
Answer: The parametric equations for the line of intersection are .
4.65. Relationships Between Planes (Tutorial 6, Example 5)

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

Click to see the solution

Let’s find the normal vectors: , , , .

  1. Planes 1 and 3:
    • . The normal vectors are parallel.
    • Check the constant terms: The equation for plane 3 is , which simplifies to . This is the same as plane 1.
    • Result: Planes 1 and 3 are coincident.
  2. Planes 2 and 4:
    • Multiply plane 4 by 3: . The normal vector is parallel to , but the constant term is different ().
    • Result: Planes 2 and 4 are parallel (but not coincident).
  3. Planes 1 and 2:
    • .
    • 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).

Answer:

  • Coincident: (1, 3)
  • Parallel: (2, 4)
  • Perpendicular: (1, 2) and (2, 3)
  • Intersecting: All other pairs.
4.66. Angle Between a Line and a Plane (Tutorial 6, Example 6)

What is the angle between the line and the plane ?

Click to see the solution
  1. Identify Direction and Normal Vectors:
    • Line: . Direction vector .
    • Plane: Normal vector .
  2. Use the Sine Formula: The angle between a line and a plane is given by .
  3. Calculate:
    • .
    • .
    • .
    • .
  4. Find the angle:
    • .
Answer: The angle between the line and the plane is .
4.67. Nonsingular Matrix Condition (Tutorial 6, Example 7)

For which , the following matrix is nonsingular? .

Click to see the solution
  1. Condition for Nonsingularity: A matrix is nonsingular if its determinant is non-zero.
  2. Calculate the determinant:
    • .
  3. Set the determinant to not equal zero:
    • .
Answer: The matrix is nonsingular for all such that and .
4.68. Unique Solution Condition (Tutorial 6, Example 8)

For which value of , does the following linear system have a unique solution?

Click to see the solution
  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:
    • .
  3. Set the determinant to not equal zero:
    • .
    • .
Answer: The system has a unique solution for all .
4.69. Nontrivial Solution Condition (Tutorial 6, Example 9)

Assume that the following linear system has nontrivial solutions. What can we say about ?

Click to see the solution
  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:
    • .
  3. Set the determinant to zero:
    • .
Answer: For the system to have nontrivial solutions, must be equal to 1.
4.70. Matrix Rank and Nonsingularity (Tutorial 6, Example 10)

Consider the vectors . 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.
Click to see the solution
  1. Form the Matrix and Find Dependencies: Let .
    • Notice that and . Vectors and are linearly dependent on .
    • To form a nonsingular matrix, we need to select a set of linearly independent columns. Since the vectors are in , the maximum size of such a set is 3. We must exclude the dependent columns and .
    • Let’s check if the remaining vectors are linearly independent by checking the determinant.
    • .
    • Since the determinant is non-zero, 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 (). Since the vectors are in , the rank cannot be greater than 3.

Answer:

  1. To make a nonsingular matrix, columns and must be excluded.
  2. The rank of the full matrix A is 3.
4.71. RREF and Gaussian Elimination (Tutorial 6, Example 11)
  1. Find RREF of: .
  2. Solve using Gaussian elimination: .
Click to see the solution
  1. Find RREF:
    • This is REF. To get RREF: .
  2. Solve System:
    • From : . Let (free variable). Then .
    • From : .

Answer:

  1. RREF is .
  2. The solution is .
4.72. REF and RREF (Tutorial 6, Example 12)
  1. Convert to REF: .
  2. Solve using RREF: .
Click to see the solution
  1. Convert to REF:
    • Notice that and . All rows are multiples of the first.
    • .
    • Or, simplifying the first row: .
  2. Solve using RREF:
    • .
    • The solution is read directly: .

Answer:

  1. REF is .
  2. The solution is .
4.73. Linear System Analysis (Tutorial 6, Example 13)
  1. Find complete solution: .
  2. Find RREF and rank of: .
  3. Solve homogeneous system: .
Click to see the solution
  1. Complete Solution:
    • Notice all three equations are multiples of the first. The system reduces to .
    • Let and be free variables. Then .
    • Solution: .
  2. RREF and Rank:
    • All rows are multiples of the first.
    • RREF: .
    • 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 .
    • Let . Then .
    • Solution: .

Answer:

  1. .
  2. RREF is as above, Rank = 1.
  3. .
4.74. System Consistency (Tutorial 6, Example 14)

Determine consistency: .

Click to see the solution
  1. Analyze the Equations:
    • The second equation is , which simplifies to . This is consistent with the first equation.
    • The third equation is , which simplifies to .
  2. Find the Contradiction: We have two conflicting requirements for the same expression: and . Since , this is a contradiction.
Answer: The system is inconsistent (has no solution).
4.75. Angle Between Vectors (Tutorial 6, Example 15)

If and , what is the angle between and ?

Click to see the solution
  1. Use Geometric Definitions:
    • .
    • .
  2. Find the Tangent: Divide the first equation by the second.
    • .
    • .
  3. Solve for the Angle:
    • .
Answer: The angle between the vectors is .
4.76. Find an Orthogonal Vector (Tutorial 6, Example 16)

If the vectors and are orthogonal and and , calculate vector .

Click to see the solution
  1. Use Vector Triple Product: We can use the identity .
  2. Apply the Identity: Let , , .
    • .
  3. Use Given Information:
    • We are given that and are orthogonal, so .
    • The equation simplifies to .
  4. Solve for :
    • .
  5. Calculate:
    • .
    • .
    • .
Answer: .
4.77. Vector Dot Product Calculation (Tutorial 6, Example 17)

If and are vectors with properties and , then what is ?

Click to see the solution
  1. Find :
    • .
    • .
    • .
    • .
  2. Expand the Target Expression:
    • .
  3. Substitute the Known Values:
    • .
Answer: The result is -2.
4.78. Find Parametric Equations for a Line (Tutorial 6, Task 1)

Find the parametric equations for the line through the points and .

Click to see the solution
  1. Find the Direction Vector: The direction vector of the line is the vector from one point to the other.
    • .
  2. Choose a Point on the Line: We can use either of the given points. Let’s choose .
  3. Write the Parametric Equations: The parametric equations of a line are given by , , , where is a point on the line and is the direction vector.
Answer: The parametric equations are , , .
4.79. Distance from Origin to a Line (Tutorial 6, Task 2)

Find the distance from the origin to the line through the point parallel to the vector .

Click to see the solution
  1. Identify the Given Information:
    • The point to find the distance from is .
    • A point on the line is .
    • The direction vector of the line is . We can use a simpler, parallel vector by dividing by 2: .
  2. Use the Distance Formula: The distance from a point to a line is .
  3. Find the Vector :
    • .
  4. Calculate the Cross Product :
    • .
  5. Calculate the Magnitudes:
    • .
    • .
  6. Calculate the Distance:
    • .
Answer: The distance from the origin to the line is .
4.80. Distance Between Two Lines (Tutorial 6, Task 3)

Determine the distance between the two lines given by:

Click to see the solution
  1. Check for Parallelism:
    • Direction vector of : .
    • Direction vector of : .
    • Since , the lines are parallel.
  2. Use the Distance Formula for Parallel Lines: The distance between two parallel lines is , where is a point on the first line, is a point on the second line, and is the common direction vector.
  3. Identify Points and the Direction Vector:
    • From , a point is .
    • From , a point is .
    • The direction vector is .
  4. Find the Vector :
    • .
  5. Calculate the Cross Product :
    • .
  6. Calculate the Magnitudes:
    • .
    • .
  7. Calculate the Distance:
    • .
Answer: The distance between the two lines is .
4.81. Find the Equation of a Plane Through Three Points (Tutorial 6, Task 4)

Find the equation of the plane through the points , , and .

Click to see the solution
  1. Find Two Vectors in the Plane: Create two vectors from the given points.
    • .
    • .
  2. Find the Normal Vector: The normal vector to the plane is perpendicular to both and . We find it by taking their cross product.
    • .
  3. Use the Point-Normal Form: Use the normal vector and one of the points, say .
    • .
  4. Simplify to the General Form:
    • .
    • .
Answer: The equation of the plane is .
4.82. Find the Equation of a Plane Containing a Point and a Line (Tutorial 6, Task 5)

Determine the equation of the plane that contains the point and the line .

Click to see the solution
  1. Find the Direction Vector and a Point on the Line:
    • Rewrite the line’s equation in symmetric form: .
    • The direction vector of the line is . To avoid fractions, we can use a parallel vector .
    • A point on the line is found by setting the expressions to 0, which gives .
  2. Find Two Vectors in the Plane:
    • One vector in the plane is the direction vector of the line, .
    • Another vector in the plane is the vector from the point on the line to the given point . Let’s call this .
  3. Find the Normal Vector: The normal vector is the cross product of the two vectors in the plane.
    • .
  4. Use the Point-Normal Form: Use the normal vector and a point in the plane, like the origin .
    • .
    • . We can divide by -3 for a simpler form.
    • .
Answer: The equation of the plane is .
4.83. Determine if Four Points are Coplanar (Tutorial 6, Task 6)

Are the four points , , , and coplanar? Why?

Click to see the solution
  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 . Let’s create vectors starting from :
    • .
    • .
    • .
  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
    • .
  4. Conclusion: Since the scalar triple product is zero, the three vectors lie on the same plane. Therefore, the four original points are coplanar.
Answer: Yes, the four points are coplanar because the scalar triple product of the three vectors formed by them is zero.
4.84. Triangle Area and Parallelepiped Volume (Midterm Recap, N1)

Given the points , , , and , find the area of triangle and the volume of the parallelepiped formed by the vectors , , and .

Click to see the solution
  1. Define the vectors originating from point A:
    • .
    • .
    • .
  2. Calculate the Area of Triangle ABC:
    • The area is half the magnitude of the cross product of two side vectors: Area .
    • First, compute the cross product: .
    • Next, find the magnitude: .
    • The area is .
  3. Calculate the Volume of the Parallelepiped:
    • The volume is the absolute value of the scalar triple product: .
    • Using the cross product from the previous step: .
    • .

Answer:

  • The area of triangle ABC is .
  • The volume of the parallelepiped is .
4.85. Linear Independence of Vectors (Midterm Recap, N2)

Determine if the vectors , , and are linearly independent.

Click to see the solution
  1. Condition for Linear Independence: Three vectors in 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).
    • .
  3. Calculate the Determinant:
    • .
  4. Conclusion: Since the determinant is (which is non-zero), the vectors are linearly independent.
Answer: Yes, the vectors are linearly independent.
4.86. Distance from a Point to a Line (Midterm Recap, N3)

Find the distance from the point to the line passing through points and .

Click to see the solution
  1. Find the Direction Vector of the Line:
    • .
  2. Find a Vector from the Line to the Point:
    • .
  3. Use the Distance Formula: The distance is given by .
  4. Calculate the Cross Product:
    • .
  5. Calculate the Magnitudes:
    • .
    • .
  6. Calculate the Distance:
    • .
Answer: The distance is .
4.87. Find Matrix Inverse using Adjugate Method (Midterm Recap, N4)

Given the matrix , find its inverse using the adjugate method.

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  1. Use the Formula: The inverse is given by .
  2. Calculate the Determinant:
    • .
  3. Find the Matrix of Cofactors ():
    • , , .
    • , , .
    • , , .
    • .
  4. Find the Adjugate Matrix: The adjugate is the transpose of the cofactor matrix, .
    • .
  5. Calculate the Inverse:
    • .
Answer: .
4.88. Find Values for a Nonsingular Matrix (Midterm Recap, N5)

For what values of is the matrix nonsingular?

Click to see the solution
  1. Condition for Nonsingularity: A matrix is nonsingular if and only if its determinant is non-zero.
  2. Calculate the Determinant:
    • .
  3. Find Values for Singularity: First, find the values of that make the matrix singular by setting the determinant to zero.
    • .
  4. Solve the Quadratic Equation:
    • Using the quadratic formula: .
    • The solutions are and .
  5. Conclusion: The matrix is singular for and . Therefore, it is nonsingular for all other values.
Answer: The matrix is nonsingular for all real numbers except and .
4.89. Solve a System of Linear Equations (Midterm Recap, N6)

Solve the system of linear equations:

Click to see the solution
  1. Write the Augmented Matrix:
  2. Use Gaussian Elimination:
    • and :
    • :
  3. Perform Back Substitution:
    • From the third row: .
    • From the second row: .
    • From the first row: .
Answer: The solution is .
4.90. Find Symmetric Equation of a Perpendicular Line (Midterm Recap, N7)

Find the symmetric equation of the line passing through point and perpendicular to the plane .

Click to see the solution
  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 .
  2. Use the Point and Direction Vector:
    • Point on the line: .
    • Direction vector: .
  3. Write the Symmetric Equation: The formula is .
    • .
Answer: The symmetric equation of the line is .
4.91. Solve a System using Cramer’s Rule (Midterm Recap, N8)

Solve the system of linear equations using Cramer’s rule:

Click to see the solution
  1. Calculate the Main Determinant (D):
    • .
  2. Calculate the Determinant for x ():
    • .
  3. Calculate the Determinant for y ():
    • .
  4. Calculate the Determinant for z ():
    • .
  5. Find the Solution:
    • .
    • .
    • .
Answer: The solution is .
4.92. Find Null Space and Column Space (Midterm Recap, N9)

Find the null space and column space of the matrix .

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  1. Find the Row Echelon Form:
    • Perform row operations: and .
    • .
  2. Find the Null Space: Solve the system , which simplifies to the single equation .
    • The first variable, , is the pivot variable. The other two, and , are free variables. Let and .
    • .
    • The general solution is .
    • 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.

Answer:

  • Null Space: The plane spanned by the vectors .
  • Column Space: The line spanned by the vector .
4.93. Analyze a System with a Parameter (Midterm Recap, N10)

For what values of does the system have a unique solution, infinitely many solutions, or no solution?

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  1. Row Reduce the Augmented Matrix:
    • and :
    • :
  2. Analyze the Last Row: The equation for the last row is .
    • Factor the expression: .
  3. Determine the Conditions:
    • No solution: This occurs when the last row is of the form where . This happens when , i.e., and .
    • Infinitely many solutions: This occurs when the last row is . This happens when , i.e., or .
    • 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 .

Answer:

  • Unique solution: Never
  • Infinitely many solutions: or
  • No solution: and
4.94. Line and Plane Intersection (Midterm, Task 1a)

Find intersection point between line and plane .

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

  1. Write line in parametric form:

    So: , ,

  2. Substitute into plane equation:

  3. Find coordinates:

Answer:

4.95. Plane Perpendicular to Line (Midterm, Task 1b)

Find the equation of a plane perpendicular to line through point .

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

  1. Identify direction vector:

    From the line:

  2. Use point-normal form:

  3. Simplify:

Answer:

4.96. Line Perpendicular to Plane (Midterm, Task 1c)

Find parametric equations for the line through perpendicular to plane .

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

  1. Identify normal vector:

    From the plane:

  2. Write parametric equations:

Answer: , ,

4.97. Line Relationship Classification (Midterm, Task 1d)

Determine the relationship between:

  • Line 1:
  • Line 2:
Click to see the solution

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

  1. Identify direction vectors:

    and

  2. Check if parallel:

    for any scalar , so not parallel.

  3. Check if intersecting:

    Set up equations: , ,

    From first: From second: From third:

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

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

4.98. Bisector Planes (Midterm, Task 1e)

Find the bisector planes of and .

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

  1. Distance formula:

    For a point :

  2. Set equal:

  3. Case 1:

  4. Case 2:

Answer: The bisector planes are and .

4.99. Scalar Triple Product and Linear Independence (Midterm, Task 2a)

Given , , , compute the scalar triple product and determine if the vectors are linearly independent.

Click to see the solution

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

  1. Calculate scalar triple product:

  2. Check linear independence:

    Since the scalar triple product , the vectors are linearly independent.

Answer: Scalar triple product = ; vectors are linearly independent.

4.100. Triangle Area Using Cross Product (Midterm, Task 2b)

Find the area of triangle with vertices , , .

Click to see the solution

Key Concept: Area =

  1. Find vectors:

  2. Calculate cross product:

  3. Find magnitude:

  4. Calculate area:

Answer:

4.101. Vector Triple Product Identity (Midterm, Task 2c)

Verify the identity for , , .

Click to see the solution

Key Concept: Compute both sides independently and verify equality.

  1. Left side - calculate :

  2. Left side - calculate :

  3. Right side - calculate dot products:

  4. Right side - final calculation:

Answer: Both sides equal . Identity verified. ✓

4.102. Matrix Rank (Midterm, Task 3a)

Find the rank of matrix .

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

  1. Row reduce:

  2. Count non-zero rows:

    There are 2 non-zero rows.

Answer: Rank =

4.103. System Solution by Gaussian Elimination (Midterm, Task 3b-i)

Solve the system using Gaussian elimination:

Click to see the solution

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

  1. Write augmented matrix:

  2. Row reduce:

  3. Back substitution:

Answer:

4.104. System Solution by Gauss-Jordan (Midterm, Task 3b-ii)

Solve the same system using Gauss-Jordan elimination.

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Key Concept: Continue from REF to RREF.

  1. Start from REF:

  2. Eliminate above pivots:

Answer:

4.105. System Solution by Cramer’s Rule (Midterm, Task 3b-iii)

Solve the same system using Cramer’s rule.

Click to see the solution

Key Concept: Calculate determinants , , , .

  1. Calculate main determinant:

  2. Calculate :

  3. Calculate :

  4. Calculate :

  5. Find solution:

Answer:

4.106. Angle Between Planes (Midterm, Task 4a)

Find the angle between planes with normal vectors and .

Click to see the solution

Key Concept: Use .

  1. Calculate dot product:

  2. Calculate magnitudes:

  3. Find angle:

Answer:

4.107. Distance from Point to Plane (Midterm, Task 4b)

Find the distance from point to plane .

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Key Concept: Use distance formula .

  1. Apply formula:

Answer:

4.108. Unique Solution Condition (Midterm, Task 4c)

For what values of does the system have a unique solution?

Click to see the solution

Key Concept: System has unique solution when .

  1. Calculate determinant:

  2. Find condition:

    For unique solution:

Answer:

4.109. Matrix Inverse (Midterm, Task 4d)

Find the inverse of using Gauss-Jordan method.

Click to see the solution

Key Concept: Augment with identity matrix, reduce to .

  1. Set up augmented matrix:

  2. Row reduce:

    Continue to get RREF…

Answer:

4.110. Determinant of AᵀA and Parallelogram Area (Homework 6, Task 1)

Let and be two vectors in , and consider the parallelogram spanned by these vectors. Define the matrix by stacking and as columns: . Prove that the area of the parallelogram is given by: Area .

Click to see the solution
  1. Area from Cross Product: The area of a parallelogram spanned by two vectors and is given by the magnitude of their cross product: Area .
  2. Lagrange’s Identity: Squaring the area gives Area. By Lagrange’s identity, the squared magnitude of the cross product can be expressed using dot products: .
  3. Compute : Now let’s compute the determinant of . First, we find the product :
  4. Relate to Dot Products: We can recognize the elements of this matrix as dot products:
    • .
  5. Calculate the Determinant: Now, we compute the determinant of this matrix:
    • .
  6. Conclusion: The expression for is identical to the expression for Area from Lagrange’s identity in step 2.
    • Therefore, Area.
    • Taking the square root gives Area .
Answer: The proof is complete.