Can we show the direction of $P'Q'$ is independent of $P$?

Can we show the direction of $P'Q'$ is independent of $P$?

Question

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 $r$.

Show that $\pi'$ is the reflection of $\pi$ 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 $L$.

Suggestion

If $\pi$ and $\pi'$ are parallel, what must be true about the normal of either plane?

How could we use the vector and scalar products here?

If $P$ is the point $(x,y,z)$, what are the coordinates of the reflection of $P$ in the origin?

If $P$ is on $L$, can we find the coordinates of $P$ in terms of a parameter $\alpha$?

Can we now find the coordinates of $P'$ and $Q'$ in terms of $\alpha$?

Solution

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!

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.

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 $\alpha$.