Planes and vectors

Problem

The plane \(E: 2x_1+x_2+2x_3=6\) as well as the points \(P(1|0|2)\) and \(Q(5|2|6)\) are given.

1. Show that the line passing through the points \(P\) and \(Q\) is perpendicular to the plane \(E\).

2. The points \(P\) and \(Q\) are symmetric about the plane \(F\). Determine an equation for \(F\).

Solution of part a

The normal vector \(\vec n\) of the plane \(E\) can be read off the plane’s equation as

\[\begin{split}\vec n = \begin{pmatrix}2\\1\\2\end{pmatrix}\end{split}\]

We choose the direction vector

\[\begin{split}\vec{PQ} = \vec Q - \vec P = \begin{pmatrix}5\\2\\6\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}4\\2\\4\end{pmatrix}\end{split}\]

for the line \(PQ\). It can be easily verfied that \(\vec{PQ}=2\vec n\) is true. The vectors \(\vec{PQ}\) and \(\vec n\) are thus collinear and therefore the line \(PQ\) is perpendicular to the plane \(E\).

We can check this graphically with Sage:

Solution of part b

Since the points \(P\) and \(Q\) are supposed to be symmetric about the plane \(F\), the line \(PQ\) is perpendicular to this plane. Furthermore, we have seen in part a that \(PQ\) is also perpendicular to plane \(E\). Thus, \(E\) and \(F\) are parallel and have the same normal vector \(\vec n\).

We choose the midpoint of the line \([PQ]\) as our reference point

\[\begin{split}\vec A = \frac{1}{2}\cdot\left(\vec P + \vec Q\right) = \frac{1}{2}\cdot\left(\begin{pmatrix}1\\0\\2\end{pmatrix}+\begin{pmatrix}5\\2\\6\end{pmatrix}\right) = \begin{pmatrix}3\\1\\4\end{pmatrix}\end{split}\]

on the plane \(F\). Its equation

\[\left(\vec X - \vec A\right)\cdot \vec n = 0\]

thus reads

\[2 x_1 + x_2 + 2 x_3 - 15 = 0\,.\]

This can again be verified by means of a 3D graphic: