Equations of Planes — Question 10

Question 10 – Solution

Strategy Expand $\|\vec r-\vec a\|^2=\|\vec r-\vec b\|^2$ and cancel $\vec r\cdot\vec r$.

General proof Expansion gives $$2(\vec b-\vec a)\cdot\vec r=\|\vec b\|^2-\|\vec a\|^2.$$ This is a plane with normal $\vec b-\vec a$. Substituting $(\vec a+\vec b)/2$ verifies that the midpoint lies on it.

Application Here $\vec b-\vec a=\left\langle 4,2,-6\right\rangle$ and the midpoint is $(3,-1,1)$. Thus $$4(x-3)+2(y+1)-6(z-1)=0,$$ or $\boxed{2x+y-3z=2}$.

Verification Direct substitution of the midpoint gives $6-1-3=2$.