The line \(L\) has equation
$$\mathbf{r} = \begin{pmatrix} 13 \\ -3 \\ -8 \end{pmatrix} + t \begin{pmatrix} -5 \\ 3 \\ 4 \end{pmatrix}$$
The point \(P\) has position vector \(\begin{pmatrix} -7 \\ 2 \\ 7 \end{pmatrix}\).
The point \(P'\) is the reflection of \(P\) in \(L\).
- Find the position vector of \(P'\).
[6]
- Show that the point \(A\) with position vector \(\begin{pmatrix} -7 \\ 9 \\ 8 \end{pmatrix}\) lies on \(L\).
[1]
- Show that angle \(PAP' = 120°\).
[3]
% Figure 3 shows kite APBP' with angle at A = 120°
\includegraphics{figure_3}
Figure 3
The point \(B\) lies on \(L\) and \(APBP'\) forms a kite as shown in Figure 3.
The area of the kite is \(50\sqrt{3}\)
- Find the position vector of the point \(B\).
[5]
- Show that angle \(BPA = 90°\).
[2]
The circle \(C\) passes through the points \(A\), \(P\), \(P'\) and \(B\).
- Find the position vector of the centre of \(C\).
[2]
[Total 19 marks]