The plane \(\pi\) has equation \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}+ \mu \mathbf{c}\) and the plane \(\pi'\) has equation \(\mathbf{r} = \mathbf{d}+ \lambda'\mathbf{e}+\mu'\mathbf{f}\), where \(\lambda\), \(\mu\), \(\lambda'\) and \(\mu'\) are scalar parameters. It is given that
\[\mathbf{a}=\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}, \mathbf{b}=\begin{pmatrix}2\\ -1 \\ -1\end{pmatrix}, \mathbf{c} = \begin{pmatrix}-1 \\ 2 \\ 1\end{pmatrix}, \mathbf{d} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}, \mathbf{e} = \begin{pmatrix}p \\ q \\ r\end{pmatrix}, \mathbf{f} = \begin{pmatrix}q\\ r\\ p\end{pmatrix},\]
\(\big \vert\mathbf{e}\big \vert=\big \vert\mathbf{f}\big \vert = \sqrt{30}\) and that \(p\) is negative. The planes \(\pi\) and \(\pi'\) are parallel. Find \(p\), \(q\) and \(r\).
We can find a vector perpendicular to \(\pi\) using the cross product \(\mathbf{b} \times \mathbf{c}= \begin{pmatrix}1\\ -1 \\ 3\end{pmatrix}\).
Since the two planes are parallel this is also perpendicular to \(\pi'\). We can use this fact in the dot product with the two direction vectors \(\mathbf{e}\) and \(\mathbf{f}\) in the plane \(\pi'\).
So \(\mathbf{e} . (\mathbf{b} \times \mathbf{c})=0\) and \(\mathbf{f} . (\mathbf{b} \times \mathbf{c})=0\).
Giving us \[\begin{align} p-q+3r &=0 \\ q-r+3p &=0 \end{align}\]Adding gives \(4p+2r=0 \quad \implies r=-2p\) and substituting in the second gives \(q=-5p\).
We now know that \(\mathbf{e}= t\begin{pmatrix}1\\-5 \\ -2\end{pmatrix}\). We also know \(\big \vert\mathbf{e}\big \vert = \sqrt{30}\) and \(p\) is negative, hence \(t=-1\) and we have
\(\mathbf{e}= \begin{pmatrix}-1\\5 \\ 2\end{pmatrix}\) and \(\mathbf{f}= \begin{pmatrix}5\\2 \\-1\end{pmatrix}\).
It might be tempting to proceed with vectors perpendicular to \(\pi\) and \(\pi'\) being \(\mathbf{b} \times \mathbf{c}\) and \(\mathbf{e} \times \mathbf{f}\).
\[\mathbf{b}\times \mathbf{c} = \left \vert \begin{matrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&-1&-1 \\ -1&2&1\end{matrix}\right \vert = \mathbf{i}-\mathbf{j}+3\mathbf{k}.\]
\[\mathbf{e}\times \mathbf{f}= \left \vert \begin{matrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ p&q&r \\ q&r&p\end{matrix}\right \vert = (qp-r^2)\mathbf{i}+(rq-p^2)\mathbf{j}+(pr-q^2)\mathbf{k}.\]
However, the resulting set of simultaneous equations is very difficult to solve.
By adding them we can then simplify using \(\big \vert\mathbf{e}\big \vert=\big \vert\mathbf{f}\big \vert = \sqrt{30}\) in other words \(p^2+q^2+r^2=30\). We can think about \((p+q+r)^2\) to continue the manipulation. Try it if you like!
Show that \(\pi'\) is the reflection of \(\pi\) in the origin.
Using the form of the plane equation \(\mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}\) with \(\mathbf{n}\) as the normal vector we found earlier, we have the equation of \(\pi\) as \[x - y + 3z = -2.\] Similarly using the point \(\mathbf{d}\), the plane \(\pi'\) is \[x - y + 3z = 2.\]
If \((x_1,y_1,z_1)\) is a point on \(\pi\) then \(x_1 - y_1 + 3z_1 = -2\). Multiplying by \(-1\) gives \((-x_1) -(-y_1) + 3(-z_1) = 2\). So the point \((-x_1,-y_1,-z_1)\) lies on \(\pi'\). Since every point on \(\pi\) has its image reflected in the origin lying on \(\pi'\), the two planes must be reflections in the origin.
The line \(L\) has equation \[\dfrac{x+1}{9}=\dfrac{y-1}{-6}=\dfrac{z}{-5}\] and lies in \(\pi\). The point \(P\) is in \(\pi\) with parameters \(\lambda = l\) and \(\mu = m\), and \(P'\) is in \(\pi'\) with \(\lambda' = l\) and \(\mu' = m\). The point \(Q'\) is the reflection of \(P\) in the origin. Show that when \(P\) lies on \(L\) the direction of \(P'Q'\) is independent of the position of \(P\) on \(L\).
Let \(\mathbf{p}\), \(\mathbf{p}'\) and \(\mathbf{q}'\) be the position vectors of \(P\), \(P'\) and \(Q'\) respectively.
We can write the vector equation of the line \(L\) by putting \(\dfrac{x+1}{9}=\dfrac{y-1}{-6}=\dfrac{z}{-5}=\alpha\), and getting \(\mathbf{r} = \begin{pmatrix}9\alpha-1\\-6\alpha+1\\-5\alpha\end{pmatrix}\). So if \(P\) is on \(L\) we have \(\mathbf{p} = \begin{pmatrix}9\alpha-1\\-6\alpha+1\\-5\alpha\end{pmatrix}\) for some \(\alpha\).
We are also told that \(P\) is on \(\pi\) with parameters \(l\) and \(m\), so we have \(\mathbf{p} = \begin{pmatrix}1+2l-m\\-l+2m\\-1-l+m\end{pmatrix}\). This gives us three equations for \(l\), \(m\) and \(\alpha\) which solve to give \[l = 4\alpha -1\text{, and } m = -\alpha.\]
Similarly we have \(\mathbf{p'}= \begin{pmatrix}-l+5m\\1+5l+2m\\1+2l-m\end{pmatrix}\), since \(P'\) is on \(\pi'\) with parameters \(l\) and \(m\) and \(\mathbf{q'}\) is \(\begin{pmatrix}-1-2l+m\\l-2m\\1+l-m\end{pmatrix}\), since \(\mathbf{q'}=-\mathbf{p}\).
Substituting in for \(l\) and \(m\), we find \(\mathbf{p'} = \begin{pmatrix}1-9\alpha\\-4+18\alpha\\-1+9\alpha\end{pmatrix}\), and \(\mathbf{q'}\) is \(\begin{pmatrix}1-9\alpha\\-1+6\alpha\\5\alpha\end{pmatrix}\).
Thus \(\mathbf{q'}-\mathbf{p'}= \begin{pmatrix}0\\-3+12\alpha\\-1+4\alpha\end{pmatrix} = (4\alpha-1)\begin{pmatrix}0\\3\\1\end{pmatrix}\), which is constant in direction whatever the value of \(\alpha\).
