Find a point on a given plane that is equally distant from three given points

Question

I have the following exercise:

Find a point on the plane $\pi:2x−3y−z+1=0$ that is equally distant from the points $A(2,3,0)$, $B(5,2,0)$, $C(1,0,0)$?

Now, I'm not sure how to do this, this is what I have thought of doing until now:

1. Find the centroid of the three points, which is the average of their coordinates: $$x=\dfrac{2+5+1}{3} = \dfrac{8}{3}, y=\dfrac{3+2+0}{3} = \dfrac{5}{3}, z=0$$

2. The three points are all lying on the $xy$ plane, whose perpendicular vector is $(0,0,1)$

3. Find the intersection between a perpendicular line to the $xy$ plane (that passes through the centroid) and the plane $\pi$.

Now, I have successfully calculated the centroid and determined the perpendicular vector to the plane determined by the three points. However, I don't know how to do the third part. Can anyone help (provided my approach is correct)?

Answer 1

One approach:

Fact: Given two points in three dimensions, the set of points equidistant to those two points is the plane that is the perpendicular bisector of the segment between the points.

In your case, you can compute three of these planes and then intersect them with the given plane. This will give you your point.

Second approach:

Fact: The circumcenter is the point that is equidistant from all the vertices of a triangle. In general, the circumcenter is not the centroid.

Fact: Three general points determine a plane, i.e., the unique plane containing them.

The set of points equidistant from three points is the line through the circumcenter of their triangle that is perpendicular to the plane determined by these points.

In your particular case, once you find the circumcenter, you've already shown how to turn this into a line. Then, all you need to do is to intersect that line with the given plane.

Since all of your points lie in the $(x,y)$-plane, your line will have the form $(x_0,y_0,t)$, where $x_0$ and $y_0$ are fixed values and $t$ can be anything. You'll just need to plug this into the equation for the plane and solve for $t$ to get your point.

Answer 2

Let $M$ be the midpoint of the line segment joining $A$ to $B$; that is, $M=\left(\frac72,\frac52,0\right)$. Now consider the line $l$ contained in $\pi$ which is the intersection of $\pi$ with the plane passing through $M$ which is orthogonal to the segment joining $A$ to $B$; that's the line$$\left\{\begin{array}{l}3x-y=8\\2x-3y-z=-1.\end{array}\right.$$Now, let $M'$ be the midpoint of the line segment joining $A$ to $C$; that is, $M'=\left(\frac32,\frac32,0\right)$. Now consider the line $l'$ contained in $\pi$ which is the intersection of $\pi$ with the plane passing through $M'$ which is orthogonal to the segment joining $A$ to $C$; that's the line$$\left\{\begin{array}{l}x+3y=6\\2x-3y-z=-1.\end{array}\right.$$The point that you are interested in is the point at which $l$ and $l'$ intersect. In order to get that point, you solve the system$$\left\{\begin{array}{l}3x-y=8\\x+3y=6\\2x-3y-z=-1.\end{array}\right.$$You will get the point $(3,1,4)$.