Finding the Vector Equation for the Intersection of Two Planes

[coolusername]

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.

Homework Equations

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.

 

[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 $t\langle 3,?,?\rangle$.

 

[pasmith]

The line of intersection between two planes obviously lies in both planes, and is therefore perpendicular to the normal of each plane. Thus the direction of the line is given by
$$\begin{pmatrix} 4 \\ 3 \\ -3 \end{pmatrix} \times \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ ? \\ ? \end{pmatrix}$$