Edexcel AEA 2011 June — Question 6

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2011
SessionJune
TopicVectors: Cross Product & Distances

6.The line \(L\) has equation $$\mathbf { r } = \left( \begin{array} { r } 13
- 3
- 8 \end{array} \right) + t \left( \begin{array} { r } - 5
3
4 \end{array} \right)$$ The point \(P\) has position vector \(\left( \begin{array} { r } - 7
2
7 \end{array} \right)\) .
The point \(P ^ { \prime }\) is the reflection of \(P\) in \(L\) .
(a)Find the position vector of \(P ^ { \prime }\) .
(b)Show that the point \(A\) with position vector \(\left( \begin{array} { r } - 7
9
8 \end{array} \right)\) lies on \(L\) .
(c)Show that angle \(P A P ^ { \prime } = 120 ^ { \circ }\) . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{12d3f92f-8464-4ba1-93a2-c7b841e3d3de-5_483_1367_1263_347} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The point \(B\) lies on \(L\) and \(A P B P ^ { \prime }\) forms a kite as shown in Figure 3.
The area of the kite is \(50 \sqrt { } 3\)
(d)Find the position vector of the point \(B\) .
(e)Show that angle \(B P A = 90 ^ { \circ }\) . The circle \(C\) passes through the points \(A , P , P ^ { \prime }\) and \(B\) .
(f)Find the position vector of the centre of \(C\) .