2020 VJC P2 Q2

2020 VJC P2 Q2

13 marks
Prelims

The points \(A\) and \(B\) have positions vectors \(\mathbf{i}-2\mathbf{j}-\mathbf{k}\) and \(s\mathbf{i}+t\mathbf{j}-4\mathbf{k}\) respectively. The plane \(p\) has equation \(3x-2y-z=5\). Given that the line \(AB\) has no points in common with the plane \(p\), express \(s\) in terms of \(t\).[2]

Find the shortest distance between the line \(AB\) and the plane \(p\).[2]

Another plane \(q\) is perpendicular to plane \(p\) and contains the line with equation \(\frac{x-5}{3}=\frac{1-y}{5}=\frac{1-z}{2}\). Find the equation of plane \(q\) in cartesian form.[3]

Find the position vector of the point which is a reflection of \(A\) in the line of intersection between \(p\) and \(q\).[6]

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Answer:\({s = \frac{2}{3}(t + 2)};\; {x - 3y + 9z = 11};\; \begin{pmatrix} 0 \\ -2 \\ 2 \end{pmatrix}\)

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