vector/plane proof

**duaneg37** · February 15th 2011, 11:47 AM

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!

**Ackbeet** · February 15th 2011, 11:50 AM

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.

**duaneg37** · February 15th 2011, 12:33 PM

$n \cdot (x-x_{0})=0$
$n=\langle a, b, c\rangle$
$\langle a, b, c\rangle \cdot (x-x_{0})=0$
$\langle a,b,c\rangle\langle x,y,z\rangle-\langle a,b,c\rangle\langle x_{0},y_{0},z_{0}\rangle=0$
$ax+by+cz-ax_{0}-by_{0}-cz_{0}=0$
$ax+by+cz+d=0$

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!

**HallsofIvy** · February 16th 2011, 05:33 AM

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 $<a, b, c>$ (the tangent plane to that vector containing $(x_0, y_0, z_0)$ ) must satisfy
$<a, b, c>\cdot<x- x_0, y-y_0, z-z_0>= 0$ .

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