(i)[2]
Find $\overrightarrow{OA} \cdot \overrightarrow{OB}$.
(ii)[1]
Hence show that there are no real values of $p$ for which $OA$ and $OB$ are perpendicular to each other.
(iii)[4]
Find the values of $p$ for which the angle $AOB = 60^\circ$.
Mathematics 9709 · AS & A Level · Coordinate geometry
Coordinate geometry — practice question
With origin $O$, point $A$ has position vector $4\mathbf{i} + 7\mathbf{j} - p\mathbf{k}$, while point $B$ has position vector $8\mathbf{i} - \mathbf{j} - p\mathbf{k}$, where $p$ is a constant.
Worked solution & mark scheme
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “The correct scalar product is $(4\mathbf{i}+7\mathbf{j}-p\mathbf{k})\cdot(8\mathbf{i}-\mathbf{j}-p\mathbf{k})=25+p^2$” …
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