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## Homework Statement

Find the distance between the planes 4x + 5y - 3z + 4 = 0, and -8x -10y + 6z + 12 = 0.

## Homework Equations

V = (x, y, z) + Y(a, b, c)

## The Attempt at a Solution

Okay, this problem has been driving me insane. Even though it's incredibly simple, and I know there's an easy formula, I haven't been able to do it the way the lecturer showed us.

Plane 1): 4x + 5y - 3z = -4

Plane 2): -8x - 10y + 6z = -12

Point on 1) = P = (-1, 0, 0)

Let v be the vector connecting P and Q (point on P2).

v = (-1, 0, 0) + x(4, 5, -3)

Intersect v and P2.

(-1, 0, 0) + x(4, 5, -3) = -12

-1 + 4x + 5x -3x = -12

6x = -11

x = -11/6

-> Q = (-1, 0, 0) - 11/6 (4, 5, -3)

Here's the problem, "Q" doesn't seem to be on P2, which was the entire point of solving for x.

e.g. sub x,y,z = (-1 - 44/6), (-55/6), (-33/6) into P2 and you should get -12 (as I understand it), but you get ~125.

Any help appreciated.