| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: show lines are skew |
| Difficulty | Standard +0.8 Part (i) is routine direction vector calculation. Part (ii) is standard skew lines verification. Part (iii) requires setting up BC perpendicular to AB using dot product equals zero, then solving for parameter t—this involves more problem-solving than typical C4 questions and combines multiple vector concepts in a non-standard way. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks |
|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix}-3\\6\\1\end{pmatrix} - \begin{pmatrix}-4\\1\\3\end{pmatrix} = \begin{pmatrix}1\\5\\-2\end{pmatrix}\ \therefore\ \mathbf{r} = \begin{pmatrix}-4\\1\\3\end{pmatrix} + s\begin{pmatrix}1\\5\\-2\end{pmatrix}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(3 - 2s = 9 + t \quad (3)\) | B1 |
| \(2\times(1) + (3)\): \(-5 = 15 + 5t\), \(t = -4\), \(s = -1\) | M1 A1 |
| sub. (2): \(1 - 5 = -7 + 12\), not true \(\therefore\) do not intersect | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OC} = \begin{pmatrix}3+2t\\-7-3t\\9+t\end{pmatrix},\ \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}6+2t\\-13-3t\\8+t\end{pmatrix}\) | M1 A1 | |
| \(\therefore \begin{pmatrix}1\\5\\-2\end{pmatrix}\cdot\begin{pmatrix}6+2t\\-13-3t\\8+t\end{pmatrix} = 0,\quad 6+2t - 65 - 15t - 16 - 2t = 0\) | M1 A1 | |
| \(t = -5\ \therefore\ \overrightarrow{OC} = \begin{pmatrix}-7\\8\\4\end{pmatrix}\) | M1 A1 | (12) |
# Question 7:
## Part (i):
$\overrightarrow{AB} = \begin{pmatrix}-3\\6\\1\end{pmatrix} - \begin{pmatrix}-4\\1\\3\end{pmatrix} = \begin{pmatrix}1\\5\\-2\end{pmatrix}\ \therefore\ \mathbf{r} = \begin{pmatrix}-4\\1\\3\end{pmatrix} + s\begin{pmatrix}1\\5\\-2\end{pmatrix}$ | M1 A1 |
## Part (ii):
$-4 + s = 3 + 2t \quad (1)$
$1 + 5s = -7 - 3t \quad (2)$
$3 - 2s = 9 + t \quad (3)$ | B1 |
$2\times(1) + (3)$: $-5 = 15 + 5t$, $t = -4$, $s = -1$ | M1 A1 |
sub. (2): $1 - 5 = -7 + 12$, not true $\therefore$ do not intersect | A1 |
## Part (iii):
$\overrightarrow{OC} = \begin{pmatrix}3+2t\\-7-3t\\9+t\end{pmatrix},\ \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}6+2t\\-13-3t\\8+t\end{pmatrix}$ | M1 A1 |
$\therefore \begin{pmatrix}1\\5\\-2\end{pmatrix}\cdot\begin{pmatrix}6+2t\\-13-3t\\8+t\end{pmatrix} = 0,\quad 6+2t - 65 - 15t - 16 - 2t = 0$ | M1 A1 |
$t = -5\ \therefore\ \overrightarrow{OC} = \begin{pmatrix}-7\\8\\4\end{pmatrix}$ | M1 A1 | **(12)**
---7. Relative to a fixed origin, the points $A$ and $B$ have position vectors $\left( \begin{array} { c } - 4 \\ 1 \\ 3 \end{array} \right)$ and $\left( \begin{array} { c } - 3 \\ 6 \\ 1 \end{array} \right)$ respectively.\\
(i) Find a vector equation for the line $l _ { 1 }$ which passes through $A$ and $B$.
The line $l _ { 2 }$ has vector equation
$$\mathbf { r } = \left( \begin{array} { c }
3 \\
- 7 \\
9
\end{array} \right) + t \left( \begin{array} { c }
2 \\
- 3 \\
1
\end{array} \right)$$
(ii) Show that lines $l _ { 1 }$ and $l _ { 2 }$ do not intersect.\\
(iii) Find the position vector of the point $C$ on $l _ { 2 }$ such that $\angle A B C = 90 ^ { \circ }$.\\
\hfill \mbox{\textit{OCR C4 Q7 [12]}}