Projection of a point on a plane

Submitted by Structure on Sat, 03/22/2008 - 20:49.

Let consider the plane (p) of equation

$(p)\:ax+by+cz+d=0$

and a point M(u,v,w)
We look for the point $N =(x_0,y_0,z_0)$ , the projection of M on the plane.Normal of the plane is the vector (a,b,c) so line by M(u,v,w) of equations

$\frac{x-u}{a}=\frac{y-v}{b}=\frac{z-w}{c}=t$

is the line perpendicular on the plane.
so we have for a point on this line parametric equations

$\left {\begin{array}{c} x=u+at\\ y=v+bt\\ z=w+ct \end{array}$

Now we want the value of $t_0$ for with a point of this line is on the plane.

$a(u+at_0)+b(v+bt_0)+c(w+ct_0)+d=0$

$t_0=-\frac{au+bv+cw+d}{a^2+b^2+c^2}$

so we have

$\left {\begin{array}{c} x_0=u+at_0\\ y_0=v+bt_0\\ z_0=w+ct_0 \end{array}$

$x_0=u-a\frac{au+bv+cw+d}{a^2+b^2+c^2}$

$y_0=v-b\frac{au+bv+cw+d}{a^2+b^2+c^2}$

$z_0=w-c\frac{au+bv+cw+d}{a^2+b^2+c^2}$

We can find the distance from the point M(u.v.w) to the plane ax+by+cz+d=0.
The square of this distance is

$d^2(M,(p))=(u-x_0)^2+(v-y_0)^2+(w-z_0)^2=(a^2+b^2+c^2)t_0^2=\frac{(au+bv+cw+d)^2}{a^2+b^2+c^2}$

so we have

$d(M,(p))=\frac{|au+bv+cw+d|}{\sqrt{a^2+b^2+c^2}}$

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Projection of a point on a plane