Question

Find two different planes whose intersection is theline:x = 6 + 3ty = -8 - 3tz = -1 - tWrite equations for each plane in the form Ax + By + C = D.

Answer 1

Answer:

$x+y=-2$ and $3z-y=5$ are two such planes.

Step-by-step explanation:

To find the two planes whose intersection is the line

$x=6+3t\\ y=-8-3t\\ z=-1-t$

You can say that t is equal to this expression

$t=\frac{x-6}{3} =\frac{y+8}{-3}=-z-1$

Next,

$\frac{x-6}{3} =\frac{y+8}{-3}\\ \frac{y+8}{-3}=-z-1$

Then,

$x+y=-2$ and $3z-y=5$ are two such planes.

You can see in the image attached that the intersection of this planes is the line

$x=6+3t\\ y=-8-3t\\ z=-1-t$