Question

I’m really struggling with a problem at the moment. The problem is to find the equation for the plane that passes through the point (1,1,1) and is parallel to x – 3y – 2z – 4 = 0.

I have determined the direction vector as (1, -3, -2) and then I calculated the parametric equations as these:

x(t) = t + 1
y(t) = 1 - 3t
z(t) = 1 - 2t

But now I can’t figure out how to determine the equation of the plane from those equations. Any help would be greatly appreciated! Thanks in advance.

Answer 1

This is a math question, not a programming question. You should ask it on math.stackexchange.com.

But since it’s so simple, I’ll just tell you how to solve it here.

You started with the plane equation x - 3y - 2z - 4 = 0. Any plane with the equation x - 3y - 2z + C = 0 (for any real number C) is parallel to your original plane.

So if you want the equation of a parallel plane passing through (1,1,1), just plug (1,1,1) into to the equation with C and then solve for C.

1 - 3*1 - 2*1 - C = 0
C = 3 + 2 - 1 = 4

So the equation of the parallel plane is x - 3y - 2z + 4 = 0.