Let S be the plane defined by 3y 3z 0 and let T be the pla

Let S be the plane defined by 3y - 3z = 0, and let T be the plane defined by x - 3y + 4z = 4. Find the vector equation for the line where S and T intersect. [x y z] = [0 0 0] + t[0 0 0]

Solution

3y - 3z = 0

x - 3y + 4z = 4

plugging z = 0 and solving x and y

3y = 0

x - 3y = 4

on solving we get y = 0 , x = 4

direction of the line is in both planes and so it is perpendicular to both the vectors

vector is parallel to the line given by ( 0,3,-3) x ( 1,-3,4 ) = ( 3, -3 , -3 )

equation of the line is

( x, y , z) = ( 4,0,0 ) + t ( 3,-3,-3 )