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} $