SPS SPS FM Pure 2021 June — Question 3 4 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2021
SessionJune
Marks4
TopicVectors 3D & Lines
TypeShow lines intersect and find intersection point
DifficultyModerate -0.3 This is a standard Further Maths vectors question requiring routine techniques: equating parametric equations to find intersection (solving simultaneous equations), then using the scalar product formula to find the angle between direction vectors. Both parts are textbook exercises with no novel insight required, making it slightly easier than average A-level difficulty.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles

Two lines, \(l_1\) and \(l_2\), have the following equations. $$l_1: \mathbf{r} = \begin{pmatrix} -11 \\ 10 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$$ $$l_2: \mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$$ \(P\) is the point of intersection of \(l_1\) and \(l_2\).
  1. Find the position vector of \(P\). [2]
  2. Find, correct to 1 decimal place, the acute angle between \(l_1\) and \(l_2\). [2]

Two lines, $l_1$ and $l_2$, have the following equations.

$$l_1: \mathbf{r} = \begin{pmatrix} -11 \\ 10 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$$

$$l_2: \mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$$

$P$ is the point of intersection of $l_1$ and $l_2$.

\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $P$.
[2]

\item Find, correct to 1 decimal place, the acute angle between $l_1$ and $l_2$.
[2]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2021 Q3 [4]}}