# Intersection between 2 planes

## Homework Statement

Find the vector equation for the line of intersection of the planes 4x+3y−3z=−5 and 4x+z=5

r = < _, _, 0> + t<3, _, _>

Fill in the blanks for the vector equation.

## The Attempt at a Solution

I used the method of elimination of linear systems.

4x + 3y - 3z = -5 (1)
4x +0y +z = 5 (2)

Subtract equation (1) from equation (2)

(1) - (2)

3y - 4z = -10

Isolate y:

y = 4z/3 -10/3

Next I isolate x from equation (2)

x = -z/4 + 5/4

Let z = t as parameter

Parametric equation:

x(t) = -t/4 + 5/4
y(t) = 4z/3 - 10/3
z(t) = t

From this I know that the point on the line of intersection is (5/4, -10/3, 0)

However for the vector of the line of intersection, I keep on getting the x value as -1/4 and the answer is 3.

## The Attempt at a Solution

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jbunniii
$z(t) = t$ isn't going to work. Try using $z(t) = ct$ where $c$ is some constant chosen to give you the 3 you need in $t\langle 3,?,?\rangle$.
$$\begin{pmatrix} 4 \\ 3 \\ -3 \end{pmatrix} \times \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ ? \\ ? \end{pmatrix}$$