Question

Find the intersection of the planes x+ (y-1)+z=0 and -x + (y+1)- z+0

Answer 1

Find the intersection of the planes $x + (y - 1) + z = 0$ and $-x + (y + 1) - z = 0$

Solution

Two planes (usually) intersect in a line. If P and Q are two points on that line, they can be thought of as vectors from the origin to the line, and an arbitrary point on the line is then given parametrically by $P + t(Q - P)$ , where t is a real parameter. If $t = 0$ , you are at point P and if $t = 1$ you are at point Q. Other values of t give other points on the line.

So what you need to do is find two points on the line.

Since the line must belong to both planes, we have a system of equations

$$\left\{ \begin{array}{l} x + (y - 1) + z = 0 \\ -x + (y + 1) - z = 0 \end{array} \right.$$

Adding first equation to second we have

$$2y = 0 \text{ and } x + z = 1$$

$P = (0,0,1)$ and $Q = (1,0,0)$ are the obvious points.

Now $Q - P = (1,0,-1)$ so the line is $(x,y,z) = (0,0,1) + t(1,0,-1)$ . Equating components gives a parametric representation of the line:

$$\begin{array}{l} x = t \\ y = 0 \\ z = 1 - t \end{array}$$

where $t$ is a real parameter.

Answer: (t,0,1-t), where t is a real parameter.