Hard +2.3 This is a challenging Further Maths question requiring students to understand how matrices transform lines in 3D space, extract direction vectors from Cartesian equations, apply matrix transformations to both a point and direction vector, then reconstruct the transformed line equation. It demands conceptual understanding of linear transformations, careful algebraic manipulation across multiple steps, and the insight that both a point on the line and its direction vector must be transformed. The non-standard presentation and requirement to 'fully justify' adds to the difficulty.
11 The Cartesian equation of the line \(L _ { 1 }\) is
$$\frac { x + 1 } { 3 } = \frac { - y + 5 } { 2 } = \frac { 2 z + 5 } { 3 }$$
The Cartesian equation of the line \(L _ { 2 }\) is
$$\frac { 2 x - 1 } { 2 } = \frac { y - 14 } { m } = \frac { z + 12 } { p }$$
The non-singular matrix \(\mathbf { N } = \left[ \begin{array} { c c c } - 0.5 & 1 & 2 \\ 1 & b & 4 \\ - 3 & - 2 & c \end{array} \right]\) maps the line \(L _ { 1 }\) onto the line \(L _ { 2 }\)
Calculate the values of the constants \(b , c , m\) and \(p\)
Fully justify your answers.
Position vector of point on \(L_1\): \(\begin{pmatrix}-1\\5\\-5/2\end{pmatrix} + \lambda\begin{pmatrix}3\\-2\\3/2\end{pmatrix}\) or \(\begin{pmatrix}-1\\5\\-5/2\end{pmatrix}+\lambda\begin{pmatrix}6\\-4\\3\end{pmatrix}\)
B1
Obtains position vector of a point on \(L_1\) or \(L_2\) (AO 2.5)
Direction vector for \(L_1\) or \(L_2\)
B1
ISW (AO 1.1b)
Position vector of point on \(L_2\): \(\begin{pmatrix}1/2\\14\\-12\end{pmatrix}+\mu\begin{pmatrix}1\\m\\p\end{pmatrix}\)
B1
Obtains vector equations for both \(L_1\) and \(L_2\) (AO 2.5)
Uses \(b\) and \(c\) to obtain value for at least one of \(m\) or \(p\) (AO 2.2a)
\(m = 2\), \(p = 4\)
A1
Obtains correct values of \(m\) and \(p\) (AO 1.1b)
# Question 11:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Position vector of point on $L_1$: $\begin{pmatrix}-1\\5\\-5/2\end{pmatrix} + \lambda\begin{pmatrix}3\\-2\\3/2\end{pmatrix}$ or $\begin{pmatrix}-1\\5\\-5/2\end{pmatrix}+\lambda\begin{pmatrix}6\\-4\\3\end{pmatrix}$ | B1 | Obtains position vector of a point on $L_1$ or $L_2$ (AO 2.5) |
| Direction vector for $L_1$ or $L_2$ | B1 | ISW (AO 1.1b) |
| Position vector of point on $L_2$: $\begin{pmatrix}1/2\\14\\-12\end{pmatrix}+\mu\begin{pmatrix}1\\m\\p\end{pmatrix}$ | B1 | Obtains vector equations for both $L_1$ and $L_2$ (AO 2.5) |
| $\begin{pmatrix}0.5+\mu\\14+m\mu\\-12+p\mu\end{pmatrix} = \begin{pmatrix}\frac{1}{2} & 1 & 2\\1 & b & 4\\-3 & -2 & c\end{pmatrix}\begin{pmatrix}-1+6\lambda\\5-4\lambda\\-2.5+3\lambda\end{pmatrix}$ | M1 | Forms matrix equation equating image of general point on $L_1$ with general point on $L_2$ (AO 3.1a); condone same parameter used twice |
| Collects and simplifies to $\begin{pmatrix}0.5-\lambda\\5b-11+\lambda(18-4b)\\-7-2.5c+\lambda(3c-10)\end{pmatrix}$ | M1 | Collects and simplifies terms (AO 1.1b) |
| $0.5 + \mu = 0.5 - \lambda \Rightarrow \mu=0, \lambda=0$; then $14 = -11+5b \Rightarrow b=5$; $-12=-7-5c/2 \Rightarrow c=2$ | M1 | Compares constant terms to obtain value for at least one of $b$ or $c$; must use different parameters (AO 3.1a) |
| $b = 5$, $c = 2$ | A1 | Obtains correct values of both $b$ and $c$ (AO 1.1b) |
| $\mu = -\lambda$, $m\mu = -2\lambda \Rightarrow m=2$; $p\mu = -4\lambda \Rightarrow p=4$ | M1 | Uses $b$ and $c$ to obtain value for at least one of $m$ or $p$ (AO 2.2a) |
| $m = 2$, $p = 4$ | A1 | Obtains correct values of $m$ and $p$ (AO 1.1b) |
11 The Cartesian equation of the line $L _ { 1 }$ is
$$\frac { x + 1 } { 3 } = \frac { - y + 5 } { 2 } = \frac { 2 z + 5 } { 3 }$$
The Cartesian equation of the line $L _ { 2 }$ is
$$\frac { 2 x - 1 } { 2 } = \frac { y - 14 } { m } = \frac { z + 12 } { p }$$
The non-singular matrix $\mathbf { N } = \left[ \begin{array} { c c c } - 0.5 & 1 & 2 \\ 1 & b & 4 \\ - 3 & - 2 & c \end{array} \right]$ maps the line $L _ { 1 }$ onto the line $L _ { 2 }$\\
Calculate the values of the constants $b , c , m$ and $p$\\
Fully justify your answers.\\
\hfill \mbox{\textit{AQA Further Paper 2 2021 Q11 [9]}}