Relative to a fixed origin \(O\), the point \(A\) has position vector \(\begin{pmatrix} -2 \\ 4 \\ 7 \end{pmatrix}\)
and the point \(B\) has position vector \(\begin{pmatrix} -1 \\ 3 \\ 8 \end{pmatrix}\)
The line \(l_1\) passes through the points \(A\) and \(B\).
- [(a)] Find the vector \(\overrightarrow{AB}\). \hfill [2]
- [(b)] Hence find a vector equation for the line \(l_1\) \hfill [1]
The point \(P\) has position vector \(\begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}\)
Given that angle \(PBA\) is \(\theta\),
- [(c)] show that \(\cos\theta = \frac{1}{3}\) \hfill [3]
The line \(l_2\) passes through the point \(P\) and is parallel to the line \(l_1\)
- [(d)] Find a vector equation for the line \(l_2\) \hfill [2]
The points \(C\) and \(D\) both lie on the line \(l_2\)
Given that \(AB = PC = DP\) and the \(x\) coordinate of \(C\) is positive,
- [(e)] find the coordinates of \(C\) and the coordinates of \(D\). \hfill [3]
- [(f)] find the exact area of the trapezium \(ABCD\), giving your answer as a simplified surd. \hfill [4]
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