Take the 2-minute tour ×  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 $$ q= \sqrt{q_0^2+q_1^2+q_2^2+q_3^2} $$  share|improve
this question thanks for editing –  user782104 Apr 21 '12 at
11:24  1 Answer 1  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