# Thread: Plane Equation

1. ## Plane Equation

Write an equation of the plane with normal vector passing through the point in scalar form:
-6x+(-7)y+(-3)z+180

This answer is incorrect. I found d to be 60, but I can't get it right - please help!

2. D is indeed 60. If you subtract 180 from 240 you get 60, and

$\displaystyle (-6)x + (-7)y + (-3)z = D$
$\displaystyle (-6)\cdot 2 + (-7)\cdot (-9) + (-3)\cdot (-3) = 60 = D$

3. If $\displaystyle n = \left\langle { - 6, - 7, - 3} \right\rangle$ is the normal the we can use any multiple, so let $\displaystyle n = \left\langle { 6, 7, 3} \right\rangle$.
The general plane is $\displaystyle 6x + 7y + 3z + d = 0$.
Now substitute the point to find d: $\displaystyle 6(2) + 7( - 9) + 3( - 3) + d = 0 \Rightarrow \quad d = 60$.

Dee, May I ask: "Do you have an instructor and/or a textbook?