Projection of a point on a plane

Posted March 22nd, 2008 by Structure

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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Comments

Errors

On September 17th, 2008 pgn674 says:

This is error ridden. After the first image, in "Normal of the plane is the vector (z,b,c)", the vector should be "(a,b,c)". In the last and second to last images, anything "bt" should be "bv". In the last image, the values of point N are expressed as x, y, and z. These letters have already been reserved for the equation of the plane.

Not errors but confusing: capitalization, spacing, and word confusion errors also occur throughout the article, causing general unreadability.

And with these fixes, I don't know whether this is actually correct. That's why I'm here; I'm trying to figure it out.

Thanks

On September 17th, 2008 Structure says:

Thanks for your observation.
We have tried to correct errors.