Question

Let S be the plane defined by x−4y−3z = −18, and let T be the plane defined by −2x+8y+7z = 41. Find the vector equation for the line where S and T intersect.
x 0 0
y = 0 + t0
z 0 0

Answer 1

1st one times 2
2x - 8y - 6z = -36
2nd
-2x + 8y + 7z= 41
add them
z = 5

sub into 1st x - 4y - 15 = -18
x - 4y = -3
x = 4y - 3
let y = 0 , x = -3, z = 5 ---> point (0,-3,5)
let y = 2 , x = 5, z = 5 ---> point (2,5,5)

so direction of line of intersection is (2, 8, 0)

using the point (2,5,5)
x = 2 + 2t
y = 5 + 8t
z = 5