A2 June 2021 Paper 2 Q11
11 The Cartesian equation of the line \(L_1\) is
\[\frac{x + 1}{3} = \frac{-y + 5}{2} = \frac{2z + 5}{3}\]The Cartesian equation of the line \(L_2\) is
\[\frac{2x - 1}{2} = \frac{y - 14}{m} = \frac{z + 12}{p}\]The non-singular matrix \(\mathbf{N} = \begin{bmatrix} -0.5 & 1 & 2 \\ 1 & b & 4 \\ -3 & -2 & c \end{bmatrix}\) 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. [9 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains a position vector of a point on \(L_1\) or \(L_2\) | B1 | 2.5 |
| Obtains a direction vector for \(L_1\) or \(L_2\), ISW | B1 | 1.1b |
| Obtains vector equations for both \(L_1\) and \(L_2\) | B1 | 2.5 |
| Forms a matrix equation equating the image of a general point on \(L_1\) with a general point on \(L_2\) Condone same parameter used twice | M1 | 3.1a |
| Collects and simplifies terms | M1 | 1.1b |
| Compares their constant terms to obtain a value for at least one of \(b\) or \(c\) Must have used different parameters for their general points | M1 | 3.1a |
| Obtains correct values of both \(b\) and \(c\) | A1 | 1.1b |
| Uses their \(b\) and \(c\) to obtain a value for at least one of \(m\) or \(p\) | M1 | 2.2a |
| Obtains correct values of \(m\) and \(p\) | A1 | 1.1b |
| (9 marks) |
Typical solution
The position vector of a point on \(L_1\) is
\[\begin{bmatrix} -1 \\ 5 \\ -5/2 \end{bmatrix} + \lambda\begin{bmatrix} 3 \\ -2 \\ 3/2 \end{bmatrix} \ \text{ or } \ \begin{bmatrix} -1 \\ 5 \\ -5/2 \end{bmatrix} + \lambda\begin{bmatrix} 6 \\ -4 \\ 3 \end{bmatrix}\]The position vector of a point on \(L_2\) is
\[\begin{bmatrix} 1/2 \\ 14 \\ -12 \end{bmatrix} + \mu\begin{bmatrix} 1 \\ m \\ p \end{bmatrix}\]\[\begin{aligned} \begin{bmatrix} 0.5 + \mu \\ 14 + m\mu \\ -12 + p\mu \end{bmatrix} &= \begin{bmatrix} -\frac{1}{2} & 1 & 2 \\ 1 & b & 4 \\ -3 & -2 & c \end{bmatrix}\begin{bmatrix} -1 + 6\lambda \\ 5 - 4\lambda \\ -2.5 + 3\lambda \end{bmatrix} \\ &= \begin{bmatrix} 0.5 - 3\lambda + 5 - 4\lambda - 5 + 6\lambda \\ -1 + 6\lambda + 5b - 4b\lambda - 10 + 12\lambda \\ 3 - 18\lambda - 10 + 8\lambda - 2.5c + 3c\lambda \end{bmatrix} \\ &= \begin{bmatrix} 0.5 - \lambda \\ 5b - 11 + \lambda(18 - 4b) \\ -7 - 2.5c + \lambda(3c - 10) \end{bmatrix} \end{aligned}\]\[0.5 + \mu = 0.5 - \lambda\]Therefore, when \(\mu = 0\), \(\lambda = 0\) so we can equate the constant terms in the equation
\[\left.\begin{aligned} 14 &= -11 + 5b \\ -12 &= -7 - 5c/2 \end{aligned}\right\} \to \begin{aligned} b &= 5 \\ c &= 2 \end{aligned}\]\[\left.\begin{aligned} \mu &= -\lambda \\ m\mu &= -2\lambda \\ p\mu &= -4\lambda \end{aligned}\right\} \to \begin{aligned} m &= 2 \\ p &= 4 \end{aligned}\]