| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | December |
| Marks | 7 |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.8 This is a Further Maths question combining 2D perpendicular distance (routine) with 3D skew line distance using the scalar triple product formula, requiring knowledge of vector methods beyond standard A-level. The 'hence' part requires geometric interpretation. While systematic, it demands techniques specific to Further Maths and multi-step vector manipulation. |
| Spec | 4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(3x + 4y = 28\) so \(a = 3, b = 4, c = 28\) (or any non-zero multiples) so \(D = \frac{ | 3 \times -6 + 4 \times 4 - 28 | }{\sqrt{3^2 + 4^2}}\) |
| \(D = 6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 4 = -\frac{4}{3}(x + 6)\) oe so | M1 | Finding equation of perpendicular line through (-6, 4) and solving simultaneously to find foot of perpendicular |
| \(-0.75x + 7 - 4 = \frac{4}{3}(x + 6) \Rightarrow x = -2.4, y = 8.8\) | A1 | |
| \(D = \sqrt{(-2.4 - -6)^2 + (8.8 - 4)^2} = 6\) | A1 | |
| (b) \(\begin{pmatrix} 2 \\ 1 \\ -4 \end{pmatrix} \times \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ -14 \\ -5 \end{pmatrix}\) | B1 | Correctly finding a mutual perpendicular BC |
| \(\left(\begin{pmatrix} 11 \\ -1 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ -2 \end{pmatrix}\right) \cdot \begin{pmatrix} -3 \\ -14 \\ -5 \end{pmatrix}\) or \(\begin{pmatrix} 7 \\ -4 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -14 \\ -5 \end{pmatrix}\) | M1 | Correct substitution into distance formula |
| \(D = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4 + 2\lambda = 11 + 3\mu, 3 + \lambda = -1 - \mu\) and \(-2 - 4\lambda = 5 + \mu\) | M1 | Looking for a PoI so all 3 (3rd might be seen later) |
| \(\lambda = -1, \mu = 1\) | A1 | |
| eg \(-2 - 4(-1) = 2 = 5 + (-3)\) so lines intersect so \(D = 0\) | A1 | Correctly solving any 2 equations. Must be checked in the unsolved equation. Value of each side must be found, not just equality asserted. |
| (c) There are two points, one on each line, such that the distance between the points is 0... | E1ft | |
| ...and so the lines must intersect. | E1ft |
**(a)** $3x + 4y = 28$ so $a = 3, b = 4, c = 28$ (or any non-zero multiples) so $D = \frac{|3 \times -6 + 4 \times 4 - 28|}{\sqrt{3^2 + 4^2}}$ | M1 | Identifying a, b and c and substituting a, b, c and ($x_1, y_1$) correctly into distance formula
$D = 6$ | A1 |
**Alternative solution:**
$y - 4 = -\frac{4}{3}(x + 6)$ oe so | M1 | Finding equation of perpendicular line through (-6, 4) and solving simultaneously to find foot of perpendicular
$-0.75x + 7 - 4 = \frac{4}{3}(x + 6) \Rightarrow x = -2.4, y = 8.8$ | A1 |
$D = \sqrt{(-2.4 - -6)^2 + (8.8 - 4)^2} = 6$ | A1 |
**(b)** $\begin{pmatrix} 2 \\ 1 \\ -4 \end{pmatrix} \times \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ -14 \\ -5 \end{pmatrix}$ | B1 | Correctly finding a mutual perpendicular BC
$\left(\begin{pmatrix} 11 \\ -1 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ -2 \end{pmatrix}\right) \cdot \begin{pmatrix} -3 \\ -14 \\ -5 \end{pmatrix}$ or $\begin{pmatrix} 7 \\ -4 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -14 \\ -5 \end{pmatrix}$ | M1 | Correct substitution into distance formula
$D = 0$ | A1 |
**Alternative solution:**
$4 + 2\lambda = 11 + 3\mu, 3 + \lambda = -1 - \mu$ and $-2 - 4\lambda = 5 + \mu$ | M1 | Looking for a PoI so all 3 (3rd might be seen later)
$\lambda = -1, \mu = 1$ | A1 |
eg $-2 - 4(-1) = 2 = 5 + (-3)$ so lines intersect so $D = 0$ | A1 | Correctly solving any 2 equations. Must be checked in the unsolved equation. Value of each side must be found, not just equality asserted.
**(c)** There are two points, one on each line, such that the distance between the points is 0... | E1ft |
...and so the lines must intersect. | E1ft |
---\begin{enumerate}[label=(\alph*)]
\item Find the shortest distance between the point $(-6, 4)$ and the line $y = -0.75x + 7$. [2]
\end{enumerate}
Two lines, $l_1$ and $l_2$, are given by
$$l_1: \mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -4 \end{pmatrix} \text{ and } l_2: \mathbf{r} = \begin{pmatrix} 11 \\ -1 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the shortest distance between $l_1$ and $l_2$. [3]
\item Hence determine the geometrical arrangement of $l_1$ and $l_2$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q5 [7]}}