# Finding the plane equation and point distance from a plane

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!

• To get $D$, you should take the absolute value which is $\sqrt{66}$ Apr 5, 2021 at 19:47
• @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 Apr 5, 2021 at 19:53
• You are correct. You could simply take $\alpha=1$. Apr 5, 2021 at 19:56

## 1 Answer

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.