# vector/plane proof

• Feb 15th 2011, 11:47 AM
duaneg37
vector/plane proof
Prove that if $a, b, c$ are not all zero, then the equation: $ax+by+cz+d=0$ represents a plane and $\langle a,b,c\rangle$ is a normal vector to the plane.

Hint if a $\neq$0, then $ax+by+cz+d=0$ is equivalent to $a(x+\frac{d}{d})+b(y-0)+c(z-0)=0$

My Calc III teacher loves proofs, but I'm not real good with them. How can I prove this? I'm not even sure how to approach it. Can anyone help? Thanks!
• Feb 15th 2011, 11:50 AM
Ackbeet
Well, the usual equation for a plane looks like this:

$\mathbf{n}\cdot(\mathbf{x}-\mathbf{x}_{0})=0,$

where $\mathbf{n}$ is a unit vector normal to the surface, $\mathbf{x}$ is a vector to the arbitrary point on the plane, and $\mathbf{x}_{0}$ is a vector to any particular point on the plane.

Can you get your equation to look like mine? If so, you're done.
• Feb 15th 2011, 12:33 PM
duaneg37
$n \cdot (x-x_{0})=0$
$n=\langle a, b, c\rangle$
$\langle a, b, c\rangle \cdot (x-x_{0})=0$
$\langle a,b,c\rangle\langle x,y,z\rangle-\langle a,b,c\rangle\langle x_{0},y_{0},z_{0}\rangle=0$
$ax+by+cz-ax_{0}-by_{0}-cz_{0}=0$
$ax+by+cz+d=0$

I know that $-ax_{0}-by_{0}-cz_{0}=d$, but is this enough to prove it? Where did you get the usual equation for a plane from? Thanks!
• Feb 16th 2011, 05:33 AM
HallsofIvy
It is simply using the fact that two vectors are perpendicular if and only if their dot product is 0. Let $(x_0, y_0, z_0)$ be any fixed point and $(x, y, z)$ be any other point. Then the vector connecting them is [tex]<x- x_0, y- y_0, z- z_0>[tex] and the "set of all points $(x, y, z)$ such that the vector [tex]<x- x_0, y- y_0, z- z_0>[tex] is perpendicular to $$ (the tangent plane to that vector containing $(x_0, y_0, z_0)$) must satisfy
$\cdot= 0$.