Help with Equation of Planes??

Question

2) Find an equation of the plane which passes through the point (5, -2, -4) that is parallel to the plane -3x - 2y - 7z = 7.

Leave your answer in the form Ax plus By plus Cz = D, where A, B, C are integers.

So for Q2 entering points into -3x - 2y - 2z=D should be -3*(5) -2*(-2) -7(-4) and equal -15 plus 4 plus 28 = 17. Does that mean -3x - 2y - 2z=17???

what happen to D = 7 in original equation??

Answer 1

The desired plane is parallel to the given plane so they have the same normal vector n.

n = <-3, -2, -7>

With the normal vector n, of the plane and a point in the plane P(5,-2,-4) we can write the equation of the plane.  Remember the normal vector of the plane is orthogonal to any vector that lies in the plane.  And the dot product of orthogonal vectors is zero.  Define R(x,y,z) to be an arbitrary point in the plane.  Then vector PR lies in the plane.

n • PR = 0

n • <R - P> = 0

<-3, -2, -7> • <x - 5, y + 2, z + 4> = 0

-3(x - 5) - 2(y + 2) - 7(z + 4) = 0

-3x + 15 - 2y - 4 - 7z - 28 = 0

-3x - 2y - 7z - 17 = 0

-3x - 2y - 7z = 17