Edexcel AEA 2013 June — Question 3 13 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2013
SessionJune
Marks13
PaperDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyChallenging +1.2 This is a multi-part vectors question requiring line intersection, angle calculation, and angle bisector. Parts (a)-(c) are standard A-level Further Maths techniques (equating parametric equations, dot product for angles). Part (d) on the angle bisector requires knowing that the bisector direction is proportional to the sum of unit direction vectors, which is slightly beyond routine but still a known result. The computational work is moderate and the question follows a guided structure typical of AEA, making it above average but not exceptionally challenging.
Spec4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point

3.The lines \(L _ { 1 }\) and \(L _ { 2 }\) have equations given by \(L _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { c } - 7 \\ 7 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { c } 2 \\ 0 \\ - 3 \end{array} \right)\) and \(L _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { c } 7 \\ p \\ - 6 \end{array} \right) + \mu \left( \begin{array} { c } 10 \\ - 4 \\ - 1 \end{array} \right)\) where \(\lambda\) and \(\mu\) are parameters and \(p\) is a constant.
The two lines intersect at the point \(C\) .
  1. Find
    1. the value of \(p\) ,
    2. the position vector of \(C\) .
  2. Show that the point \(B\) with position vector \(\left( \begin{array} { c } - 13 \\ 11 \\ - 4 \end{array} \right)\) lies on \(L _ { 2 }\) . The point \(A\) with position vector \(\left( \begin{array} { c } - 7 \\ 7 \\ 1 \end{array} \right)\) lies on \(L _ { 1 }\) .
  3. Find \(\cos ( \angle A C B )\) ,giving your answer as an exact fraction. The line \(L _ { 3 }\) bisects the angle \(A C B\) .
  4. Find a vector equation of \(L _ { 3 }\) .

3.The lines $L _ { 1 }$ and $L _ { 2 }$ have equations given by\\
$L _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { c } - 7 \\ 7 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { c } 2 \\ 0 \\ - 3 \end{array} \right)$ and $L _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { c } 7 \\ p \\ - 6 \end{array} \right) + \mu \left( \begin{array} { c } 10 \\ - 4 \\ - 1 \end{array} \right)$\\
where $\lambda$ and $\mu$ are parameters and $p$ is a constant.\\
The two lines intersect at the point $C$ .
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the value of $p$ ,
\item the position vector of $C$ .
\end{enumerate}\item Show that the point $B$ with position vector $\left( \begin{array} { c } - 13 \\ 11 \\ - 4 \end{array} \right)$ lies on $L _ { 2 }$ .

The point $A$ with position vector $\left( \begin{array} { c } - 7 \\ 7 \\ 1 \end{array} \right)$ lies on $L _ { 1 }$ .
\item Find $\cos ( \angle A C B )$ ,giving your answer as an exact fraction.

The line $L _ { 3 }$ bisects the angle $A C B$ .
\item Find a vector equation of $L _ { 3 }$ .
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2013 Q3 [13]}}