A2 June 2019 Paper 2 Q2
2
Find the shortest distance between \(\Pi\) and \(C\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\left|\begin{pmatrix} 4 \\ -5 \\ 1 \end{pmatrix}.\begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix} - 15\right|}{\left|\begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix}\right|}\) | M1 | 1.1a |
| \(\left|\begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix}\right| = \sqrt{3^2 + 6^2 + 2^2}\) or \(\begin{pmatrix} 4 \\ -5 \\ 1 \end{pmatrix}.\begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix} = 4 \times 3 + (-5) \times 6 + 1 \times (-2)\) | M1 | 1.1 |
| 5 | A1 | 1.1 |
| [3] |
Notes
M1: Substitution into formula. Condone missing mod signs on top.
Numerator can be \(\left|\mathbf{q}.\begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix}\right|\) where \(\mathbf{q}\) is vector from C to point on \(\Pi\).
Or substituting \(\mathbf{r} = \begin{pmatrix} 4 \\ -5 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix}\) into equation of \(\Pi\) to find a value for \(\lambda\).
M1: Soi
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}\) | M1 | |
| \(\begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}.\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \sqrt{11}\sqrt{6}\cos\theta\) | *M1 | |
| \(\cos\theta = \pm\dfrac{6}{\sqrt{66}}\) | A1 | |
| \(d = |\mathbf{r}|\sin\theta = |\mathbf{r}|\sqrt{1 - \cos^2\theta}\) | dep*M1 | |
| \(\sqrt{5}\) | A1 | |
| [5] |
Notes
M1: Finding the vector \(\mathbf{r}\) from a point on one line to a point on the other
*M1: Using dot product to find an expression for \(\cos\theta\). NB Ignore attempts to use the formula for the distance between 2 skew lines.
A1: can be implied by \(\cos\theta = 0.739\), or \(\theta = 0.740\) or 2.402 or \(42.4^\circ\) or \(137.6^\circ\)
Alternate method 1
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}\) | M1 |
| \(\left(\begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}\right).\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 0\) | *M1 |
| \((\mu = -1\) so\()\ \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix}\) | A1 |
| \(d = \sqrt{(4 - 4)^2 + (3 - 4)^2 + (1 - 3)^2}\) | dep*M1 |
| \(\sqrt{5}\) | A1 |
| [5] |
M1: Finding the vector from a particular point on one line to a general point on the other
*M1: Dotting the vector found with one of the direction vectors and setting to 0
A1: Solving to find (a parameter value and hence) position vector or coordinates of equivalent point on other line, or (\(\lambda = -\frac{1}{2}\) and) \(\begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix}\)
dep*M1: Distance formula for their 2 points
Alternate method 2
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}\) | M1 |
| \(|\mathbf{r}| = \sqrt{6\mu^2 + 12\mu + 11}\) | *M1 A1 |
| \(\mu = -1\) | dep*M1 |
| \(d = \sqrt{5}\) | A1 |
| [5] |
M1: Finding the vector from a particular point on one line to a general point on the other. As above
dep*M1: Attempts to find parameter (Calculus or completes square)
Alternate method 3
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}\) | M1 |
| \(\left|\begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\right| = \left|\begin{pmatrix} 5 \\ 2 \\ -1 \end{pmatrix}\right|\) | *M1 |
| \(\ldots = \sqrt{5^2 + 2^2 + 1^2} = \sqrt{30}\) | A1 |
| \(d = \dfrac{\sqrt{30}}{\left|\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\right|} = \sqrt{5}\) | dep*M1 A1 |
| [5] |
M1: Finding the vector \(\mathbf{r}\) from a point on one line to a point on the other
*M1: Calculates cross product of ‘their \(\mathbf{r}\)’ and direction vector. Allow if full method seen or if 2 terms correct