Given two points$ P_{1}(0, 0, 0)$ and $P_{2}(2, 2, 0)$ what is the plane equation equidistant from $P_1$ and $P_2$?

Question

I have given two points $P_{1}(0, 0, 0)$ and $P_{2}(2, 2, 0)$ what is the plane equation equidistant from $P_1$ and $P_2$?

How can I find this?

Answer 1

by the distance formula, $$(x-2)^2+(y-2)^2+(z-0)^2 = x^2+y^2+z^2$$ Which gives $x+y=2$.

Alternatively, $(2-0,2-0,0-0)$ is direction ratio for the normal to the plane.The midpoint $M(1,1,0)$ must lie on the plane. Let $P(x, y, z)$ be any other point on the plane. The the direction ratio of $PM(x-1, y-1, z-0)$ is perpendicular to above mentioned normal. Take their dot product and put it zero$$(2,2,0)•(x-1, y-1, z) =0$$ giving $$2(x-1)+2(y-1)=0$$which is $$x+y=2$$

Answer 2

If point $P$ is equidistance from $P_1$ and $P_2$ then this can expressed as follows:

$ (P - P_1) \cdot (P - P_1) = (P - P_2) \cdot (P - P_2) $

Expanding and simplifying,

$ P \cdot (P_2 - P_1) = \dfrac{1}{2} (P_2 \cdot P_2 - P_1 \cdot P_1 ) $

The right side can also be written as

$ \dfrac{1}{2} (P_2 - P_1) \cdot (P_2 + P_1) $

Putting it all together

$ P \cdot (P_2 - P_1) = \dfrac{1}{2} (P_2 - P_1) \cdot (P_2 + P_1) $

Which can re-written as

$ \left(P - \dfrac{1}{2} (P_1 + P_2) \right) \cdot (P_2 - P_1) = 0 $

And this is an equation of a plane whose normal vector is $(P_2 - P_1)$ and which passes through the point $\dfrac{1}{2} (P_1 + P_2) $.