Relative to the origin $O$, the position vectors of points $A$ and $B$ are $\overrightarrow{OA} = i + 2j + 2k$ and $\overrightarrow{OB} = 3i + 4j$. Point $P$ is on the straight line through $A$ and $B$, with $\overrightarrow{AP} = \lambda \overrightarrow{AB}$.
(i)[2]
Show that, after simplification, $\overrightarrow{OP} = (1 + 2\lambda)i + (2 + 2\lambda)j + (2 - 2\lambda)k$.
(ii)[5]
By comparing the expressions for $\cos AOP$ and $\cos BOP$ in terms of $\lambda$, find the value of $\lambda$ for which $OP$ bisects $\angle AOB$.
(iii)[1]
When $\lambda$ takes this value, check that $AP : PB = OA : OB$.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Apply a correct method to write $\overrightarrow{OP}$ in terms of $\lambda$” …