Closest point on a plane to a point near the plane

[kairama15]

TL;DR Summary

Trying to find the point on a plane that is closest to another point near the plane.

Suppose there is a 3d plane z=a*x+b*y+c.

Suppose there is a point in space near, but not on the plane. (xo, yo, zo).

What is the coordinate (x1,y1,z1) on the plane that is nearest the original point?

My attempt uses minimization but the result is blowing up into large answer. I wonder if there is an easier way to solve this that gives an elegant general solution. Can anyone help with this?

Thanks!

 

[fresh_42]

Do you know what a normal vector to a plane is and how to calculate it?

 

[jedishrfu]

Consider the point to be a vector from the origin to the point call it ##P_0##

Consider the plane's normal vector ##N_0## and the point you want to find as ##P_1##

Ask yourself what multiple of the normal vector when added to the outside point vector will intersect the plane.

 

[kairama15]

Yes.

The normal vector will be
<-a,-b,1> right?

I imagine this unit vector is orthogonal to the plane and should pass through both points being considered, right?

 

[mfb]

Right.

 

[kairama15]

So if i know that the gradient vector is <-a,-b,1>, and i want that vector to start at (xo,yo,zo) and move towards the plane, I can set up parametric equatioms to describe this movement through 3d space.

x=xo-a*t
y=yo-b*t
z=zo+(1)*t

Since I want to determine where this parametric point crosses through the plane, I can plug each 'function of t' x y and z into x y and z in the plane's equation z=ax+by+c.

I get:
zo+t=a*(xo-a*t)+b*(yo-b*t) + c

Solving for t:
t= (a*xo+b*yo+c-zo)/(1+a^2+b^2).

I can than plug t into the parametric equations to find the point on the plane that is closest to the point (xo,yo,zo).

Example for new x coordinate:
x1 = xo - a*(a*xo+b*yo+c-zo)/(1+a^2+b^2)

Is this line of reasoning correct, and is my math correct? Thanks everyone.

 

[fresh_42]

The idea is correct. I haven't checked your formulas. If you have a point (outside the plane) and a direction (normal vector), then you get a straight which intersects with the plane at exactly one point. This means the point we are looking for is the point which satisfies both equations (straight and plane).