Review question

# Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5089

## Solution

The lines $l_1$ and $l_2$ have equations $\mathbf{r} = \mathbf{a} + s\mathbf{b}$ and $\mathbf{r} = \mathbf{c} + t\mathbf{d}$, where $s$ and $t$ are real numbers and

$\mathbf{a} = 5\mathbf{i}+ 6\mathbf{j}- \mathbf{k}, \quad\mathbf{b} = \mathbf{i} +\mathbf{j}+ \mathbf{k}, \quad\mathbf{c} = \mathbf{i}+ 13\mathbf{j} +7\mathbf{k}, \quad\mathbf{d} = \mathbf{i} + 6\mathbf{j} + 3\mathbf{k}.$

Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order:

1. show that $PQ = \sqrt{38},$

2. find the position vectors of $P$ and $Q,$

We can write the position vector of $P$ as $\mathbf{p} = \begin{pmatrix}5+s \\ 6+s \\ -1+s\end{pmatrix}$, and the position vector of $Q$ as $\mathbf{q} = \begin{pmatrix}1+t \\ 13+6t\\ 7+3t\end{pmatrix}.$

Thus $\mathbf{q} - \mathbf{p} = \begin{pmatrix}t-s-4 \\ 6t-s+7 \\ 3t-s+8\end{pmatrix}$. This vector has to be perpendicular to both $l_1$ and $l_2$, and so using the scalar product, $\begin{pmatrix}t-s-4 \\ 6t-s+7 \\ 3t-s+8\end{pmatrix}.\begin{pmatrix}1 \\ 1\\1\end{pmatrix}=0 \quad\text{and}\quad \begin{pmatrix}t-s-4 \\ 6t-s+7 \\ 3t-s+8\end{pmatrix}.\begin{pmatrix}1\\ 6 \\ 3\end{pmatrix}=0.$

This gives us the two equations $10t-3s+11=0$ and $46t-10s+62=0$, which solve to give $s = -3$, $t = -2$.

Thus $\mathbf{p} = \begin{pmatrix}2 \\ 3 \\ -4\end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix}-1 \\ 1\\ 1\end{pmatrix}$. We also have $\mathbf{q}-\mathbf{p} = \begin{pmatrix}-3 \\ -2\\ 5\end{pmatrix}$, and so $\big\vert\mathbf{q}-\mathbf{p} \big\vert = \sqrt{(-3)^2+(-2)^2+5^2} = \sqrt{38}$.

1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$.

Using vector product we can find a vector perpendicular to two others, thus $(\mathbf{q}- \mathbf{p}) \times \mathbf{b}=\mathbf{n}$ where $\mathbf{n}$ is a vector perpendicular to both $\mathbf{b}$ (the direction vector of $l_1$) and $PQ$.

$\begin{pmatrix}-3\\-2\\5\end{pmatrix} \times \begin{pmatrix}1 \\ 1\\1\end{pmatrix}=\begin{pmatrix}-2-5\\ -(-3-5)\\-3--2\end{pmatrix}=\begin{pmatrix}-7 \\ 8\\-1\end{pmatrix}$

A normal vector immediately yields the vector equation of the plane $\mathbf{r}.\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane.

So $\mathbf{q}.\mathbf{n}= 7+8-1=d=14$.

Hence the vector equation of the plane is $\begin{pmatrix}x\\y\\z\end{pmatrix} . \begin{pmatrix} -7 \\ 8\\-1\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$

Check; is $\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes.

As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\mathbf{r} = \begin{pmatrix}x \\ y\\ z\end{pmatrix}= \begin{pmatrix}2\\ 3 \\ -4\end{pmatrix}+ \lambda\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} + \mu\begin{pmatrix}3 \\ 2\\ -5\end{pmatrix}.$

Thus $x = 2+\lambda + 3\mu$, $y = 3+\lambda + 2\mu$, $z = -4+\lambda -5\mu$.

Subtracting the first two, $x-y=-1+\mu$, and so $\mu = x-y+1$.

Subtracting the second two, $y-z= 7+7\mu$, and so $y-z-7=7(x-y+1)$.

This becomes $7x-8y+z+14=0$, which is the equation of the plane we seek.