Find the equation of the plane that contains all the points that are equidistant from the given points
$(-9, 3, 3), (6, -2, 4)$
I think the plane described lies in the midpoint of these points, and it is perpendicular to the line connecting the two points. This means that the point $\left(\dfrac{15}{2}, \dfrac{1}{2}, \dfrac{7}{2}\right)$ is on the plane, and vector line perpendicular to the plane is $<-9-6, 3-(-2), 3-4>=<-15, 5, -1>$ . So the equation of the plane is $$-15(x-\dfrac{15}{2})+5(y-\dfrac12)-(z-\dfrac72)=0$$ or $$15x-5y+z-\dfrac{227}{2}=0$$ However, this seems to be the wrong answer. What am I doing wrong?
Best Answer
The midpoint is $\left(\frac{-3}{2},\frac{1}{2},\frac{7}{2} \right)$