Vectors: Question 5

Worked solution

Part (a): Vector equation of line $PQ$

$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix}6\\3\\-3\end{pmatrix} - \begin{pmatrix}2\\-1\\5\end{pmatrix} = \begin{pmatrix}4\\4\\-8\end{pmatrix}$

Every component of $\begin{pmatrix}4\\4\\-8\end{pmatrix}$ is divisible by $4$, so the simplest integer direction vector is $\begin{pmatrix}1\\1\\-2\end{pmatrix}$ (checking: $4\times\begin{pmatrix}1\\1\\-2\end{pmatrix}=\begin{pmatrix}4\\4\\-8\end{pmatrix}$ ✓).

Using $P$ as the fixed point, a vector equation of the line is:

$\mathbf{r} = \begin{pmatrix}2\\-1\\5\end{pmatrix} + t\begin{pmatrix}1\\1\\-2\end{pmatrix}$

Part (b): Does $R(-2,-5,13)$ lie on line $PQ$?

Setting $\mathbf{r} = \begin{pmatrix}-2\\-5\\13\end{pmatrix}$ and solving each component for $t$:

• $x$: $2+t=-2 \implies t=-4$
• $y$: $-1+t=-5 \implies t=-4$
• $z$: $5-2t=13 \implies -2t=8 \implies t=-4$

All three components independently give $t=-4$. Because the value of $t$ is consistent across $x$, $y$ and $z$, the point $R$ does lie on line $PQ$ (at $t=-4$).

Check by recomputing in reverse order (starting from $z$): $5-2t=13 \implies t = \dfrac{5-13}{2} = \dfrac{-8}{2}=-4$; then $y$: $-1+(-4)=-5$ ✓; then $x$: $2+(-4)=-2$ ✓. Same value of $t$ found again, confirming $R$ lies on $PQ$.

Part (c): Unit vector in the direction of $\overrightarrow{PQ}$

Using the simplified direction vector $\begin{pmatrix}1\\1\\-2\end{pmatrix}$, its magnitude is:

$\left|\begin{pmatrix}1\\1\\-2\end{pmatrix}\right| = \sqrt{1^2+1^2+(-2)^2} = \sqrt{1+1+4} = \sqrt6$

So the unit vector is:

$\frac{1}{\sqrt6}\begin{pmatrix}1\\1\\-2\end{pmatrix} = \frac{1}{\sqrt6}\mathbf{i} + \frac{1}{\sqrt6}\mathbf{j} - \frac{2}{\sqrt6}\mathbf{k}$

Check using the un-simplified vector instead: $\overrightarrow{PQ}=\begin{pmatrix}4\\4\\-8\end{pmatrix}$ has magnitude $\sqrt{4^2+4^2+(-8)^2}=\sqrt{16+16+64}=\sqrt{96}=4\sqrt6$. Dividing gives $\dfrac{1}{4\sqrt6}\begin{pmatrix}4\\4\\-8\end{pmatrix} = \dfrac{1}{\sqrt6}\begin{pmatrix}1\\1\\-2\end{pmatrix}$. The same unit vector, confirming the result is independent of which (equivalent) direction vector is normalised.

Final answers

• (a) $\mathbf{r} = \begin{pmatrix}2\\-1\\5\end{pmatrix} + t\begin{pmatrix}1\\1\\-2\end{pmatrix}$
• (b) Yes, $R$ lies on line $PQ$ (at $t=-4$)
• (c) $\dfrac{1}{\sqrt6}\mathbf{i}+\dfrac{1}{\sqrt6}\mathbf{j}-\dfrac{2}{\sqrt6}\mathbf{k}$