Find the equation of a plane containing line $L$ and equidistant from two points

Question

I want to find the equation for a plane that is equidistant from the points $P_0 = (1, 2, 3)$ and $P_1=(3, 2, 1)$ and contains the line $\ell=(1, 1, 1) + t(0, 1, -1).$

I know that the midpoint of the two points $P_0$ and $P_1$ must be on the plane, which is $MP=(2, 2, 2).$ However, I notice this point is on the line $\ell$. I need another vector in the plane to preform the cross product and find a normal to the plane, however I can't think of a way to find such vector. I know that the length of the orthogonal complements of the two points on the normal must be the same for all points on the plane, but setting up such an equation doesn't seem like the right way to go about it. Any help would be appreciated.

Answer 1

The set of points that is equidistant from two points in $\mathbb R^3$ is a plane. The normal vector of this plane is $\overrightarrow{P_0P_1}$ and contains the midpoint $(2,2,2)$.

Therefore the place you require is

$$2x-2z = 0$$

or

$$x=z$$

However this plane does not contain the line $l$; their intersection is the point $(1,1,1)$, so I suspect something is wrong with the question: the line $l$ itself is not equidistant from the two points.

Answer 2

The plane that is equidistant from the points $P_0=(1,2,3);\;P_1=(3,2,1)$ is the symmetry plane of the segment $P_0P_1$. Its equation can be found calling $P=(x,y,z)$ a generic point of the space and solving $$PP_0^2=PP_1^2$$ $$(x-1)^2+(z-3)^2=(x-3)^2+(z-1)^2$$ That simplifies in $$x-z=0$$ which does not contain the line $\ell=(1,1+t,1-t)$ because in the parametric equation of $\ell$ we have $x\ne z$.

Answer 3

There are actually two planes satisfying the given condition.

I use the following notations:

• $n=\begin{pmatrix}n_x \\ n_y\\ n_z\end{pmatrix}$ a normal vector of a plane satisfying the given condition
• $d=\begin{pmatrix}0 \\ 1\\ -1\end{pmatrix}$ the direction vector of line $l$
• $P_l =\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$ the position vector of point $P_l(1,1,1)$ on line $l$

Then we must have

$$n\cdot d = 0 \Rightarrow n_y = n_z$$

Furthermore, because of the equal distances of points $P_0$ and $P_1$ to the plane, we have

$$\left|(P_0-P_l)\cdot n\right| = \left|(P_1-P_l)\cdot n\right|$$ $$\Rightarrow \left|n_y+2n_z\right| = \left|2n_x+n_y\right|$$ or since $n_y = n_z$ $$3|n_y| = |2n_x+n_y|$$

Solving this, you get (up to a non-zero scalar) the normal vectors

$$n =\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} \text{ or } n =\begin{pmatrix}-2 \\ 1\\ 1\end{pmatrix}$$

Hence, the two planes are

$$x+y+z = 3 \text{ or } -2x+y+z = 0$$