AS June 2019 Paper 1 Q3
3 The position vector of point \(A\) is \(\mathbf{a} = -9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}\).
The line \(l\) passes through \(A\) and is perpendicular to \(\mathbf{a}\).
\(l\) is also perpendicular to the vector \(\mathbf{b}\) where \(\mathbf{b} = -2\mathbf{i} + \mathbf{j} + \mathbf{k}\).
\(P\) is a point on \(l\) such that \(PA = 2OA\).
\(C\) is a point whose position vector, \(\mathbf{c}\), is given by \(\mathbf{c} = p\mathbf{a}\) for some constant \(p\). The line \(m\) passes through \(C\) and has equation \(\mathbf{r} = \mathbf{c} + \mu\mathbf{b}\). The point with position vector \(9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k}\) lies on \(m\).
| Scheme | Marks | AO |
|---|---|---|
| e.g. Shortest distance is the length of the perpendicular from \(O\) to \(l\), so length \(OA\). | B1 | 2.2a |
| \(\sqrt{\mathbf{a}.\mathbf{a}} = \sqrt{(-9)^2 + 2^2 + 6^2} = 11\) | B1 | 1.1 |
| [2] |
Notes
B1: (1st) Can be implied by attempting to find length \(OA\)
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{a} \times \mathbf{b} = (-9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} + \mathbf{k}) = )\) \(-4\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\) | B1 | 1.1 |
| [1] |
Notes
B1: BC. Or any non-zero multiple
Allow column vectors
| Scheme | Marks | AO |
|---|---|---|
| eg \(\mathbf{r} = (-9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) + \lambda(4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k})\) | B1ft | 1.1 |
| [1] |
Notes
B1ft: Must be \(\mathbf{r} = \ldots\) or \(x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = \ldots\) oe. (must be an equation)
Allow column vectors
Allow any equivalent equation
| Scheme | Marks | AO |
|---|---|---|
| PAO is a right angled triangle | M1 | 3.1a |
| \(\tan\theta = 2\) | A1 | 1.1 |
| 1.11 rads or \(63.4^\circ\) | A1 | 1.1 |
| [3] |
Notes
M1: May be implied by diagram or attempt at trigonometric ratio which implies that there is a right angle at \(A\).
A1: (1st) Correct trigonometric equation satisfied by \(\theta\)
A1: (2nd) Note that correct answer with no working seen is full credit
SC: If trying to find vector \(\overrightarrow{OP}\)
B1 for using \(\overrightarrow{AP} = \lambda\begin{pmatrix} 4 \\ 3 \\ 5 \end{pmatrix}\) and finding that \(\lambda = \dfrac{11\sqrt{2}}{5}\) oe
B1 for correct use of cosine rule to find correct angle
B1 for answer of 1.11 (1.107..) or \(63.4^\circ\)
| Scheme | Marks | AO |
|---|---|---|
| Need \(9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k} = p\mathbf{a} + \mu\mathbf{b}\) | M1 | 2.2a |
| (any two of) \(-9p - 2\mu = 9\), \(2p + \mu = 8\), \(6p + \mu = -12\) | M1 | 1.1 |
| \(p = -5\) | A1 | 1.1 |
| [3] |
Notes
M1: (1st) Could instead consider \(p\mathbf{a} = 9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k} + \mu\mathbf{b}\)
This would give \(\mu = -18\)
M1: (2nd) Equating coefficients for two of \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\)
Only 2 equations are necessary
\(\mu\) might be the negative value
A1: BC or eliminating \(\mu\) (\(= 18\))
No need to check for consistency