Edexcel C4 — Question 7 13 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks13
PaperDownload PDF ↗
TopicVectors 3D & Lines
TypeShow lines intersect and find intersection point
DifficultyStandard +0.3 This is a standard C4 vectors question testing routine techniques: finding intersection by equating components (part a), using dot product for angle between lines (part b), calculating vector magnitudes (part c), and applying parallelogram properties (part d). All parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting

The line \(l_1\) has vector equation $$\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}$$ and the line \(l_2\) has vector equation $$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix},$$ where \(\lambda\) and \(\mu\) are parameters. The lines \(l_1\) and \(l_2\) intersect at the point \(B\) and the acute angle between \(l_1\) and \(l_2\) is \(\theta\).
  1. Find the coordinates of \(B\). [4]
  2. Find the value of \(\cos \theta\), giving your answer as a simplified fraction. [4]
The point \(A\), which lies on \(l_1\), has position vector \(\mathbf{a} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). The point \(C\), which lies on \(l_2\), has position vector \(\mathbf{c} = 5\mathbf{i} - \mathbf{j} - 2\mathbf{k}\). The point \(D\) is such that \(ABCD\) is a parallelogram.
  1. Show that \(|\overrightarrow{AB}| = |\overrightarrow{BC}|\). [3]
  2. Find the position vector of the point \(D\). [2]

The line $l_1$ has vector equation
$$\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}$$
and the line $l_2$ has vector equation
$$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix},$$
where $\lambda$ and $\mu$ are parameters.

The lines $l_1$ and $l_2$ intersect at the point $B$ and the acute angle between $l_1$ and $l_2$ is $\theta$.

\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $B$. [4]

\item Find the value of $\cos \theta$, giving your answer as a simplified fraction. [4]
\end{enumerate}

The point $A$, which lies on $l_1$, has position vector $\mathbf{a} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}$.
The point $C$, which lies on $l_2$, has position vector $\mathbf{c} = 5\mathbf{i} - \mathbf{j} - 2\mathbf{k}$.
The point $D$ is such that $ABCD$ is a parallelogram.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that $|\overrightarrow{AB}| = |\overrightarrow{BC}|$. [3]

\item Find the position vector of the point $D$. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4  Q7 [13]}}