Taking $O$ as the origin, the position vectors of $A$ and $B$ are $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$. The line $l$ is given by $\mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.
(a)[4]
Find the acute angle between the directions of $AB$ and $l$.
(b)[5]
Find the position vector of the point $P$ on $l$ for which $AP = BP$.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State or show that $\vec{AB}=(2,-1,-3)$” …