Find the plane equation which contains all of the points that are at the same distance from points $A(4,-2,2)$ and $B(-1,2,-3)$ We can say that this plane is a bisector to AB. after finding the plane equation find the distance between the point $C(10,-4,1)$ to the plane.

What I did was:

A point which will be on the plane according to the given information is the mid point so the mid point of $AB$ is $MD=(1.5,0,-0.5)$ so in order to find a plane equation I need a point ($MD$)and a normal , I don't know if I can assume that $(-5,4,-5)$ is Perpendicular to the plane (according to $l(t)=(-1,2,-3)+t(-5,4,-5)$) so after that I assumed that the plane equation is $Ax+By+Cz+D=0$ then the normal is $(A,B,C)$ and since I assumed that $(5,-4,5)$ is perpendicular to the plane it means $(A,B,C)=\alpha(-5,4,-5)$ after substituting the points we get $D=5$ so out equation is $5x-4y+5z-5=0$ after that I just applied the distance of point from plane equation and got $D=\frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}}$ , $D=\frac{|-5*10+(4*-4)-5+5}{\sqrt{66}}$ =$-\sqrt{66}$ which is close to $-8.12$ and one of the possible answers is $8.12$ but I honestly felt like I have no idea what I am doing with the question and just tried to use what I know so I don't know if what I did or my answer is even correct.

appreciate any help and tips that will help with better understanding these topics thank you!

  • $\begingroup$ To get $ D $, you should take the absolute value which is $\sqrt{66}$ $\endgroup$ Apr 5, 2021 at 19:47
  • $\begingroup$ @hamam_Abdallah yes of course , my mistake but my point is to know if the way is correct , and actually if it is correct then why is it correct? final answer is not that important $\endgroup$
    – Adamrk
    Apr 5, 2021 at 19:53
  • $\begingroup$ You are correct. You could simply take $ \alpha=1$. $\endgroup$ Apr 5, 2021 at 19:56

1 Answer 1


Your working is correct. Specifically on your point - I don't know if I can assume that (−5,4,−5) is Perpendicular to the plane

Say, the midpoint of line segment $AB$ is $M$.

As we know in $2D$, all points on perpendicular bisector of a line segment is equidistant from the ends of the line segment. As all points on the plane are equidistant from points $A$ and $B$, all lines lying in the plane and going through intersection point of $AB$ and the plane, must be perpendicular bisector of the line segment $AB$. For that to happen, the plane must be normal to $AB$ and intersect it at $M$, its midpoint.

As you obtained, point $\small M \ (1.5, 0, - 0.5)$ lies on the plane and we know that vector $(-5, 4, -5)$ is normal to the plane, we can write the equation of the plane as

$ - 5 (x-1.5) + 4(y-0) - 5 (z+0.5) = 0 \implies 5x - 4y + 5z = 5$ as you obtained and also the distance from point $C$ is $\sqrt{66}$, as you calculated.


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