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Suppose I have two points $A=(x_1,y_1,z_1)$ and $B=(x_2,y_2,z_2)$. Suppose I want to find the equation of the plane equidistant from these points of form $ax+by+cz+d=0$. What is the equation in terms of the coordinates of $A$ and $B$

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3 Answers 3

Assuming you mean that all points in the plane are equidistant from $A$ and $B$, then notice that the vector $v=\langle x_2-x_1,y_2-y_1,z_2-z_1 \rangle$ is normal to the desired plane. Furthermore, the midpoint of the line segment between $A$ and $B$ must lie in this plane. The midpoint is the point $c=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$. Can you come up with the equation for a plane if you have a normal vector and a point that lies in the plane?

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For calculating a plane you need a point and a normal vector.

A point $P$ of your plane is fore sure the center point of $A$ and $B$. (because of the same distances to $A$ and $B$. $$P = \frac{A+B}{2}$$

A normal vector $\overrightarrow{n}$ onto the plane is just the vector $\overrightarrow{AB}$. $$\overrightarrow{n}=\overrightarrow{AB} = B - A$$

Then you can creat your plane with: $$\overrightarrow{n}X=\overrightarrow{n}P$$

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If the point $P = (x,y, z)$ is on the bisector plane, then we know $distance(P,A) = distance(P,B)$. Therefore $$ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 = (x-x_2)^2 + (y-y_2)^2 + (z-z_2)^2$$

If you multiply everything out, you'll find that all the squared terms disappear, and you'll be left with an equation that can be massaged into the form $ax+by+cz+d =0$.

Alternatively, there is a geometric approach: we know that the plane passes through the point that's half way between $A$ and $B$. This is the point $(A+B)/2$.

Also we know that the normal vector of the plane is the vector between $A$ and $B$, which is $B-A$.

Then, maybe you know how to get the equation of a plane if you have its normal vector and a point that lies on it ?

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