A2 June 2025 Paper 1 Q11
11 The lines \(l_1\) and \(l_2\) have equations
\(l_1: \dfrac{x - 3}{a} = \dfrac{y + 1}{b} = \dfrac{z - 2}{1} \qquad l_2: \mathbf{r} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} + \lambda(-\mathbf{i} + c\mathbf{j} + 2\mathbf{k})\)
where \(a\), \(b\) and \(c\) are constants.
| Scheme | Marks | AO |
|---|---|---|
| \(l_1\) has direction vector \(a\mathbf{i} + b\mathbf{j} + \mathbf{k}\) | M1 | 3.1a |
| \((-\mathbf{i} + 2\mathbf{k}) \cdot (a\mathbf{i} + b\mathbf{j} + \mathbf{k}) = 0\) or \(-a + 2 = 0\) | M1 | 1.1 |
| \(\Rightarrow a = 2\) | A1 | 1.1 |
| \(2 - \lambda = 3 + 2\mu,\; 1 = -1 + \mu b,\; 3 + 2\lambda = 2 + \mu\) | M1 | 2.1 |
| \(\Rightarrow \lambda + 2\mu = -1,\; 2\lambda - \mu = -1\) \(\Rightarrow \lambda = -0.6\) or \(\mu = -0.2\) | A1 | 1.1 |
| Point of intersection is \(\left(\frac{13}{5}, 1, \frac{9}{5}\right)\) | A1 | 3.2a |
| [6] |
Notes
M1: direction vector for line \(l_1\) soi
M1: scalar product = 0
M1: equating coordinates of two lines to get two equations in \(\lambda\) and \(\mu\) or three equations in \(\lambda\), \(\mu\) and \(b\). Could still be in terms of \(a\).
A1: finding \(\lambda\) or \(\mu\)
A1: or \(x = \frac{13}{5}, y = 1, z = \frac{9}{5}\) but do not accept a position vector
| Scheme | Marks | AO |
|---|---|---|
| (i) parallel if \(a\mathbf{i} + b\mathbf{j} + \mathbf{k} = \lambda(-\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})\) | M1 | 3.1a |
| \(\lambda = 0.5\) | A1 | 1.1 |
| \(\Rightarrow a = -\tfrac{1}{2},\; b = 2\) | A1 | 1.1 |
| [3] | ||
| (ii) distance between lines \(= \dfrac{|\mathbf{u} \times \mathbf{v}|}{|\mathbf{v}|}\) where \(\mathbf{u} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} - (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) | M1* | 3.1a |
| \(\mathbf{v} = -\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) \(\mathbf{u} \times \mathbf{v} = \mathbf{j} - 2\mathbf{k}\) | M1dep A1 | 1.1 1.1 |
| \(|\mathbf{u} \times \mathbf{v}| = \sqrt{5} \qquad |\mathbf{v}| = \sqrt{21}\) | M1dep | 1.1 |
| distance \(= \sqrt{\frac{5}{21}}\) | A1 | 1.1 |
| [5] |
Notes
(b)(i)
M1: soi
A1: or one correct value of \(a\) or \(b\) (not from incorrect working)
(b)(ii)
M1*: vector between one point on each line, e.g. \(\mathbf{i} - 2\mathbf{j} - \mathbf{k}\); could use any such points. Must be seen or implied by correct vector product. Do not condone use of direction vector.
M1dep: vector product with their \(\mathbf{u}\) and a multiple of the direction vector; allow a sign slip only
A1: correct vector product for their vectors. Cannot be implied.
M1dep: modulus formula for both of their vectors
A1: oe 0.49 or better, www
Alternative method 1
| Scheme | Marks |
|---|---|
| Minimise distance between \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \((2 - \lambda)\mathbf{i} + (1 + 4\lambda)\mathbf{j} + (3 + 2\lambda)\mathbf{k}\) \((1 + \lambda)\mathbf{i} - (2 + 4\lambda)\mathbf{j} - (1 + 2\lambda)\mathbf{k}\) | M1 |
| \(|(1 + \lambda)\mathbf{i} - (2 + 4\lambda)\mathbf{j} - (1 + 2\lambda)\mathbf{k}|\) \(21\lambda^2 + 22\lambda + 6\) | A1 |
| \(42\lambda + 22 = 0\) or \(21\left(\lambda + \frac{11}{21}\right)^2 + \frac{5}{21}\) | M1 |
| \(\lambda = -\dfrac{11}{21}\) | A1 |
| distance \(= \sqrt{\frac{5}{21}}\) | A1 |
M1: considering the vector between a point on one line and a general point on the other; could use any such points.
A1: or \(\sqrt{21\lambda^2 + 22\lambda + 6}\), correct for their distance or distance squared
M1: minimising their distance or distance squared; must be seen
A1: soi by correct answer and completed square form
A1: oe 0.49 or better, www
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} -1 - t \\ 2 + 4t \\ 1 + 2t \end{pmatrix}\) | M1 |
| \(\begin{pmatrix} -1 - t \\ 2 + 4t \\ 1 + 2t \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix}\) | M1 |
| \(-(-1 - t) + 4(2 + 4t) + 2(1 + 2t) = 0\) | M1 |
| \(t = -\dfrac{11}{21}\) | A1 |
| distance \(= \sqrt{\frac{5}{21}}\) | A1 |
M1: considering the vector between a general point on one line and a general point on the other; could be in terms of any variable, e.g. \((-1 - (\lambda - \mu))\mathbf{i} + (2 + 4(\lambda - \mu))\mathbf{j} + (1 + 2(\lambda - \mu))\mathbf{k}\)
M1: considering the scalar product between this vector and any multiple of the direction vector
M1: writing scalar product as a sum and equating to zero. Soi by \(t = -\frac{11}{21}\).
A1: solving their correct equation correctly
A1: oe 0.49 or better, www
Alternative method 3
| Scheme | Marks |
|---|---|
| \(\mathbf{u} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} - (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) | M1* |
| \(\cos\theta = \dfrac{(-\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}) \cdot (-\mathbf{i} + 2\mathbf{j} + \mathbf{k})}{\sqrt{(-1)^2 + 4^2 + 2^2}\sqrt{(-1)^2 + 2^2 + 1^2}}\) | M1dep |
| \(\cos\theta = \dfrac{11}{\sqrt{6}\sqrt{21}}\) | A1 |
| distance \(= \sqrt{(-1)^2 + 2^2 + 1^2}\,\sin\theta\) | M1dep |
| distance \(= \sqrt{\frac{5}{21}}\) | A1 |
M1*: vector between one point on each line, e.g. \(\mathbf{i} - 2\mathbf{j} - \mathbf{k}\); could use any such points. Must be seen or implied by correct scalar product. Do not condone use of direction vector.
M1dep: scalar product with their \(\mathbf{u}\) and a multiple of the direction vector; allow a sign slip only. \(\cos\theta\) must be seen.
A1: implied by \(\theta = 11.49^\circ\) or \(\theta = 0.2005\) rad
M1dep: \(|\mathbf{u}|\sin\theta\) ft their values; dependent on M1M1 having been awarded
A1: oe 0.49 or better, www