Question

Find the equations of the line of intersection of the following planes. a. \(2 x-3 y+2 z=5\) and \(x+2 y-z=4\). b. \(3 x+y-2 z=1\) and \(x+y+z=5\).

Short answer

For part (a), \((x, y, z) = (2 + 8t, 1 - 5t, 7t)\). For part (b), \((x, y, z) = (-2 + 3s, 7 - 5s, 2s)\).

Step-by-step solution

Identify the Planes

The equations given represent two planes each. For part (a), the first plane equation is \(2x - 3y + 2z = 5\) and the second is \(x + 2y - z = 4\). For part (b), the first plane is \(3x + y - 2z = 1\) and the second is \(x + y + z = 5\).

Find Normal Vectors of Each Plane

The normal vector of a plane \(ax + by + cz = d\) is \(\langle a, b, c \rangle\). For part (a), normal vectors are \(\langle 2, -3, 2 \rangle\) and \(\langle 1, 2, -1 \rangle\). For part (b), normal vectors are \(\langle 3, 1, -2 \rangle\) and \(\langle 1, 1, 1 \rangle\).

Determine the Cross Product of Normal Vectors

The direction vector of the line of intersection is the cross product of the normal vectors. For part (a), compute \(\langle 2, -3, 2 \rangle \times \langle 1, 2, -1 \rangle\). For part (b), compute \(\langle 3, 1, -2 \rangle \times \langle 1, 1, 1 \rangle\).

Compute Cross Products

Use the determinant formula to find the cross product:- For part (a): \(\langle 8, -5, 7 \rangle\).- For part (b): \(\langle 3, -5, 2 \rangle\).

Solve for a Point on the Line

Choose either of the plane equations to solve for the points on the line by setting one variable to zero. For part (a), set \(z = 0\) and solve the system:- \(2x - 3y = 5\)- \(x + 2y = 4\).For part (b), set \(z = 0\):- \(3x + y = 1\)- \(x + y = 5\).

Solve Simple Systems

For part (a): Substitute to find \((x, y, z) = (2, 1, 0)\).For part (b): Substitute to find \((x, y, z) = (-2, 7, 0)\).

Write the Parametric Equations

Using the point found and the direction vector:- For part (a), the parametric equations are: \[ x = 2 + 8t, \; y = 1 - 5t, \; z = 7t \]- For part (b), the parametric equations are: \[ x = -2 + 3s, \; y = 7 - 5s, \; z = 2s \]

Key concepts

Normal Vector

A normal vector is a key component in understanding planes and their interactions. Imagine a plane as a flat surface, much like a sheet of paper. The normal vector is like an arrow pointing directly perpendicular to that sheet of paper. Mathematically, if you have a plane given by the equation \( ax + by + cz = d \), then the normal vector is \( \langle a, b, c \rangle \). This vector helps in determining how the plane is oriented in space.
In the given problem, each plane has its own normal vector based on its equation coefficients:

• For the plane \( 2x - 3y + 2z = 5 \), the normal vector is \( \langle 2, -3, 2 \rangle \).
• For the plane \( x + 2y - z = 4 \), it is \( \langle 1, 2, -1 \rangle \).
• Similarly, for the second set of planes, their normal vectors are \( \langle 3, 1, -2 \rangle \) and \( \langle 1, 1, 1 \rangle \).

Normal vectors from two planes are crucial in finding the line of intersection because they define the orientation of both planes.

Cross Product

The cross product is an essential tool when dealing with vectors. It allows us to find a vector that is perpendicular to two given vectors. In the context of our problem, the cross product is used to find the direction of the line where two planes intersect.
The cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is calculated using the determinant formula:\[ \mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3 \\end{array} \right| \]This results in a new vector \( \mathbf{c} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \). This vector \( \mathbf{c} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).
For instance, in our exercise:

• For part (a), the cross product \( \langle 2, -3, 2 \rangle \times \langle 1, 2, -1 \rangle \) results in the direction vector \( \langle 8, -5, 7 \rangle \).
• For part (b), \( \langle 3, 1, -2 \rangle \times \langle 1, 1, 1 \rangle \) results in \( \langle 3, -5, 2 \rangle \).

These direction vectors are then used to construct the equations of the line of intersection.

Parametric Equations

Parametric equations are a powerful way to describe lines in three-dimensional space. They express the coordinates \( (x, y, z) \) of points on the line as functions of a parameter, often named \( t \) or \( s \).
The basic form of parametric equations for a line is:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

Here, \( (x_0, y_0, z_0) \) is a point on the line, and \( \langle a, b, c \rangle \) is a direction vector. The parameter \( t \) varies over all real numbers.
To determine parametric equations for the line where two planes intersect:

• First, find a specific point on the line by solving a system of equations derived from the planes.
• Next, use the direction vector obtained from the cross product.

In the problem:

• For part (a), with point \( (2, 1, 0) \) and direction vector \( \langle 8, -5, 7 \rangle \), the parametric equations are:
  \( x = 2 + 8t, \; y = 1 - 5t, \; z = 7t \).
• For part (b), with point \( (-2, 7, 0) \) and direction vector \( \langle 3, -5, 2 \rangle \), they are:
  \( x = -2 + 3s, \; y = 7 - 5s, \; z = 2s \).

These equations offer a convenient representation of the line in three-dimensional space.