Find equation for all points equidistant to two points

Question

I have two points: A(-2,5,4) and B(6,2,-3). Per the question, I want to find an equation of all points equidistant from those two points. The intuition and visualization is easy, this will be a plane of points such that point C, whatever it may be, creates an isosceles triangle with points A and B. Vectors are still alien to me and I need help writing the problems in mathematical terms.

$A=(-2, 5, 4)$
$B=(6, 2, -3)$
$\vec{AB}=<8,-3,-7>=8\hat{i}-3\hat{j}-7\hat{k}$
$\hat{AB}=\frac{\vec{AB}}{\sqrt{122}}$
$Midpoint=(2, 3.5, 0.5)$

I don't know where to go from here, assuming I did this part correctly. All I know is I need a plane that passes through the midpoint and is perpendicular to $\vec{AB}$. I thought about using $\vec{AB} \cdot \vec{AB^{⊥}}=0$ and setting $\vec{AB^{⊥}}=<a,b,c>$, but WebAssign rejected my end result.

Can somebody walk me through this problem and explain why you took each step?

Answer 1

Let $\vec r$ be the position vector of any general point on this plane. Then, $\vec r - (2, \ 3.5, \ 0.5)$ will represent any general vector lying on the plane. Now, we need all such vectors to be perpendicular to the normal vector of the plane, i.e. the one you obtained $(\vec{AB})$. In other words, equate the dot product to zero: $$\big((\vec r -(2, \ 3.5, \ 0.5)\big) \cdot (8, \ -3, \ -7) =0$$ One can write this equation in cartesian form as $$8(x-2) -3(y-3.5) -7(z-0.5)=0$$

Answer 2

A plane perpendicular to $\vec{AB}=\hat{i}-3\hat{j}-7\hat{k}$ and passing through the origin has equation $$8x-3y-7z=0.\tag1$$ If you've learned about dot products, note that the dot product with $\vec{AB}$ of any vector in this plane is $0$. A plane parallel to $(1)$ has equation $$8x-3y-7z=k$$ for some constant $k$, so all that remains to do is to determine the value of $k$ such that midpoint lies on the plane.

Over to you.

Answer 3

Why not just work from the original definition of the set? The distance from an arbitrary point to $A$ is $\sqrt{(x+2)^2+(y-5)^2+(z-4)^2}$ and from $B$ is $\sqrt{(x-6)^2+(y-2)^2+(z+3)^2}$. Setting these equal to each other and squaring, we have $$(x+2)^2+(y-5)^2+(z-4)^2 = (x-6)^2+(y-2)^2+(z+3)^2.$$ The squared terms all cancel, leaving $$8x-3y-7z-2=0.$$

You’ve got a good start on a solution using vectors. The midpoint is indeed $(2,7/2,1/2)$ and a normal vector to the plane is $(8,-3,-7)$. There’s no particularly good reason to normalize this vector in order to solve this problem, though. You’re just making extra work for yourself by doing that. Now, the normal vector to the plane is orthogonal to the vector between any two point on the plane. So, if you have a normal $\mathbf n$ to the plane and a known point $\mathbf x_0$ on it, an equation of the plane is $\mathbf n\cdot(\mathbf x-\mathbf x_0)=0$. Plug in your values and rearrange as necessary. You should end up with exactly the same equation as above.