Review question

# Where are $P$ and $Q$ if $PQ$ is perpendicular to $l_1$ and $l_2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8828

## Solution

The equations of the lines $l_1$ and $l_2$ are $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$ and $\mathbf{r}=\mathbf{c}+\mu \mathbf{d}$ respectively, where $\mathbf{a}=-6\mathbf{i}+3\mathbf{j}+15\mathbf{k}, \hspace{0.5cm} \mathbf{b}=\mathbf{i}-2\mathbf{j}+3\mathbf{k}, \hspace{0.5cm} \mathbf{c}=6\mathbf{i}+15\mathbf{j}+39\mathbf{k}, \hspace{0.5cm} \mathbf{d}=2\mathbf{i}-3\mathbf{j}+4\mathbf{k},$ and where $\lambda$ and $\mu$ are scalar parameters. The points $P$ and $Q$ are on $l_1$ and $l_2$ respectively, and $PQ$ is perpendicular to both $l_1$ and $l_2$. In any order:

1. find the position vectors of $P$ and $Q$.

The question tells us we are free to answer the parts of the question in any order that we like. Here we will start with the last part, by finding the vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$.

The points $P$ and $Q$ can be represented in vector form as $p\mathbf{i}+q\mathbf{j}+r\mathbf{k}$ and $s\mathbf{i}+t\mathbf{j}+u\mathbf{k}$.

If now $P$ and $Q$ lie on $l_1$ and $l_2$ respectively, they have to satisfy the equation of each line. We can hence deduce the six equations

\begin{align*} -6+\lambda &= p,\\ 3-2\lambda &= q,\\ 15+3\lambda &= r,\\ 6+2\mu &= s,\\ 15-3\mu &= t,\\ 39+4\mu &= u. \end{align*}

But we also know that the vector corresponding to $\overrightarrow{PQ}=(p-s)\mathbf{i}+(q-t)\mathbf{j}+(r-u)\mathbf{k}$ has to be perpendicular to both lines, and hence to the two direction vectors $b$ and $d$. Two vectors are perpendicular if and only if their scalar product vanishes. Hence we have the two additional equations

\begin{align*} (p-s)-2(q-t)+3(r-u)&=0,\\ 2(p-s)-3(q-t)+4(r-u)&=0. \end{align*}

This gives us 8 equations in total for 8 unknowns. Combining the first six equations with the last two leaves us with

\begin{align*} 20\lambda&=84+29\mu,\\ 7\lambda&=30+10\mu, \end{align*}

which can be solved to give $\lambda=10$ and $\mu=4$. Substituting these values back into the equations for $l_1$ and $l_2$ yields the position vectors for our points:

\begin{align*} \overrightarrow{OP}&=4\mathbf{i}-17\mathbf{j}+45\mathbf{k},\\ \overrightarrow{OQ}&=14\mathbf{i}+3\mathbf{j}+55\mathbf{k}. \end{align*}
1. show that $PQ=10\sqrt{6}$;

The length of a vector $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ is given by $\sqrt{x^2+y^2+z^2}$. Hence all we have to do is to calculate the difference of the position vectors $P$ and $Q$ and calculate the corresponding length, giving

\begin{align*} PQ=\sqrt{(4-14)^2+(-17-3)^2+(45-55)^2}=10\sqrt{6}. \end{align*}
1. find a vector which is parallel to $\overrightarrow{PQ}$, giving your answer in the form $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$;

We already know the vector $\overrightarrow{PQ}=10\mathbf{i}+20\mathbf{j}+10\mathbf{k}$. Multiplying this by any constant gives a vector parallel to $\overrightarrow{PQ}$. For example, we can multiply by $-\frac{1}{10}$ to get $\mathbf{i}+2\mathbf{j}+\mathbf{k}$.

Tackling part (iii) before part (i) may seem strange here. A different order might have made better sense if we had used a different method. For instance, we could use the vector cross product to find a vector perpendicular to the two lines in which case it would be easiest to answer (i) first.