Solving Problems with Parallelograms, Lines, and Planes in 3D Space

Finding the Vertex Coordinates of a Parallelogram

The points A (-2, 3, 1), B (2, -1, 3), and C (0, 1, -2) are consecutive vertices of the parallelogram ABCD.

(a) Find the Vertex Coordinates of D

If ABCD are the vertices of a parallelogram, free vectors AB and DC are equal:

• AB = (4, -4, 2)
• DC = (-x, 1 - y, -2 - z)

Equating coordinates, we have x = -4, y = 5, and z = -4. The missing point is D (-4, 5, -4).

(b) Equation of the Line Through B and Parallel to Diagonal AC

The line passes through point B (2, -1, 3) and has a direction vector AC = (2, -2, -3). Its continuous equation is:

(x - 2) / 2 = (y + 1) / -2 = (z - 3) / -3

(c) Equation of the Plane Containing the Parallelogram

We can use point B (2, -1, 3) and the vectors BA = (-4, 4, -2) and BC = (-2, 2, -5) as independent parallel vectors. Its parametric equation is:

• x = 2 - 4L - 2u
• y = -1 + 4L + 2u
• z = 3 - 2L - 5u

where L and u are real numbers.

Analyzing the Relationship Between a Line and a Plane

R is the line given by:

• 2x + y - mz = 2
• x - y - z = -m

and the plane π is defined by x + my - z = 1.

A principal vector of the line is the vector product of the normal vectors of each plane:

u = | i   j   k |
        | 2   1   -m|
        | 1   -1   -1 | = i(-1 - m) - j(-2 + m) + k(-3) = (-1 - m, 2 - m, -3)

A normal vector of the plane is n = (1, m, -1).

(a) Is There a Value of m for Which π and r are Parallel?

If the line r is parallel to the plane π, the principal vector of r, u, is perpendicular to the normal vector of π, n, so their dot product is zero:

u • n = 0 = (-1 - m, 2 - m, -3) • (1, m, -1) = -m² + m + 2 = 0

Solving the equation yields m = -1 and m = 2. Therefore, for m = -1 and m = 2, the line r is parallel to the plane π.

(b) What is the Value of m When the Line is in the Plane?

If the line is contained in the plane, they must be parallel. From part (a), we know that m = -1 or m = 2. We solve the system of equations for m = -1 and m = 2. The value of m for which the system has infinite solutions (two equations and three unknowns) is the desired value.

If m = -1, the system is:

• 2x + y + z = 2
• x - y - z = 1
• x - y - z = 1

The second and third equations are identical. The system reduces to:

• 2x + y + z = 2
• x - y - z = 1

Thus, for m = -1, the line r lies in the plane π.

If m = 2, the system is:

• 2x + y - 2z = 2
• x - y - z = -2
• x + 2y - z = 1

Adding the second and third equations and then subtracting the first, we get:

0x + 0y + 0z = 3

This is a contradiction. Therefore, for m = 2, the line is parallel to the plane but not contained within it.

(c) Relative Position of the Line and Plane When m = 0

For m = 0, the line intersects the plane at a single point. We can find this point by solving the system:

• 2x + y = 2
• x - y - z = 0
• x - z = 1

Subtracting the third equation from the second, we find y = 1. Substituting into the first equation, we get x = 1/2. Substituting into the second equation, we find z = -1/2. The intersection point is (x, y, z) = (1/2, 1, -1/2).