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 $B$, and $overrightarrow{AP} = \lambda \overrightarrow{AB}$.
(i)[2]
Show, from the information above, that $\overrightarrow{OP} = (1 + 2\lambda)i + (2 + 2\lambda)j + (2 - 2\lambda)k$.
(ii)[5]
By setting the expressions for $\cos AOP$ and $\cos BOP$ equal in terms of $\lambda$, determine the value of $\lambda$ for which $OP$ bisects angle $AOB$.
(iii)[1]
Verify that, when $\lambda$ takes this value, $AP : PB = OA : OB$.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use a valid method to write $\overrightarrow{OP}$ in terms of $\lambda$” …