Question

Find an equation of the plane that passes through the line of intersection of the planes x − z = 3 and y + 3z = 3
and is perpendicular to plane
x + y − 3z = 6

Answer 1

Intersection planes are x - z = 3 and y + 3z = 3

x - z - 3 = 0 and y + 3z - 3 = 0

parametic equation of the plane equation x - z - 3 + t(y + 3z - 3) = 0 ---> (1)

x - z - 3 + yt + 3zt - 3t = 0

x + yt + z( 3t - 1) - 3t  - 3 = 0

x + yt + z(3t - 1) + (- 3t - 3) = 0

Normal vector of the plane is < 1, t, (3t - 1) >

This normal perpendicular to given plane x + y - 3z = 6.

The vector is < 1, 1, - 3 >

So, dot product between < 1, t, (3t -1) > . < 1, 1, - 3 > = 0

1(1) + t(1) + (3t - 1)(-3) = 0

1 + t - 9t + 3 = 0

8t = 4

t = 1/2

Substitute the t value in equation (1).

x - z - 3 + 1/2 (y + 3z - 3) = 0

x - z - 3 + y/2 + 3z/2 - 3/2 = 0

x + y/2 + z/2 - 9/2 = 0

Required plane equation is 2x + y + z - 9 = 0.