Relative to origin $O$, the position vectors of $A$, $B$, $C$, $D$ are $\overrightarrow{OA} = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$, $\overrightarrow{OB} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$, $\overrightarrow{OC} = 2\mathbf{i} + 4\mathbf{j} + \mathbf{k}$, $\overrightarrow{OD} = -3\mathbf{i} + \mathbf{j} + 2\mathbf{k}$.
(i)[6]
Find the plane equation through $A$, $B$ and $C$, and give your answer in the form $ax + by + cz = d$.
(ii)[5]
The line through $D$ parallel to $OA$ intersects the plane with equation $x + 2y - z = 7$ at the point $P$. Find the position vector of $P$ and show that the length of $DP$ is $2\sqrt{14}$.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Determine a vector parallel to the plane, e.g. $\vec{AB}=\mathbf{i}-2\mathbf{j}-3\mathbf{k}$” …