# The normal vector to the plane $5x+2y+3z=1$?

I am calculating the normal vector to the plane $5x+2y+3z=1$

According to the book:

The normal vector $N$ is often normalized to unit length because in that case the equation $$d = N ⋅Q + D$$ gives the signed distance from the plane to an arbitrary point $Q$. If $d = 0$, then the point $Q$ lies in the plane. If $d > 0$, we say that the point $Q$ lies on the positive side of the plane since $Q$ would be on the side in which the normal vector points.

How to get thenormal vector $N$? Thank you

Updated: the normailize function $$q= \sqrt{q_0^2+q_1^2+q_2^2+q_3^2}$$

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thanks for editing –  user782104 Apr 21 '12 at 11:24

If $N$ is the unit normal to the plane, $p$ is a point in the plane and $q$ is any point then the signed distance $d$ of $q$ to the plane is $$\langle(p-q), N\rangle = d$$ In coordinate representation this reads $$\sum_i N_i (p_i-q_i) = d$$ The equation you have describes the set of $(x,y,z)$ which make up the plane. So simply by comparison, $n := (5,2,3)^T$ is a normal vector to your plane. It's not of unit length, though, so you need to normalize it if you want to have that. That is, $$N = \frac{1}{\sqrt{38}}(5,2,3)^T$$ (because $5^2 + 2^2 + 3^2 = 38$) and the distance of your plane to the origin is $\sqrt{38}$.