Projection of a point on a plane

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

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

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

$$\left<br />
{\begin{array}{c}<br />
x_0=u+at_0\\<br />
y_0=v+bt_0\\<br />
z_0=w+ct_0<br />
\end{array}$$
$$x_0=u-a\frac{au+bv+cw+d}{a^2+b^2+c^2}$$
$$y_0=v-b\frac{au+bv+cw+d}{a^2+b^2+c^2}$$
$$z_0=w-c\frac{au+bv+cw+d}{a^2+b^2+c^2}$$

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

$$d(M,(p))=\frac{|au+bv+cw+d|}{\sqrt{a^2+b^2+c^2}}$$
Average: 4.4 (145 votes)

Back to top