| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: unknown constant then intersect |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring equating components to find intersection, then using the scalar product formula for angle between lines. All techniques are routine and well-practiced, though it involves multiple steps and careful algebraic manipulation. Slightly easier than average due to the structured guidance through parts (i)-(iii). |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(4s = 6 + 14r\) ... (1) | B1 | |
| \(-3 - 2x = 3 + 2t\) ... (2) | B1 | |
| \((1) + 2 × (2): -6 = 12 + 18t, t = -1, s = -2\) | M1 A1 | |
| \(r = \begin{pmatrix} 7 \\ 0 \\ -3 \end{pmatrix} + (-1)\begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ -8 \\ 1 \end{pmatrix}\) | A1 | |
| (ii) \(a - (-5) = -3, a = -8\) | M1 A1 | |
| (iii) \(\cos \theta = \frac{ | 5×(-5) + 4×14 + (-2)×2 | }{\sqrt{25+16+4}×\sqrt{25+196+4}}\) |
| \(= \frac{27}{\sqrt{45×15}} = \frac{9}{3\sqrt{5}×5} = \frac{3}{5\sqrt{5}} = \frac{3}{25}\sqrt{5}\) | M1 A1 |
**(i)** $4s = 6 + 14r$ ... (1) | B1
$-3 - 2x = 3 + 2t$ ... (2) | B1
$(1) + 2 × (2): -6 = 12 + 18t, t = -1, s = -2$ | M1 A1
$r = \begin{pmatrix} 7 \\ 0 \\ -3 \end{pmatrix} + (-1)\begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ -8 \\ 1 \end{pmatrix}$ | A1
**(ii)** $a - (-5) = -3, a = -8$ | M1 A1
**(iii)** $\cos \theta = \frac{|5×(-5) + 4×14 + (-2)×2|}{\sqrt{25+16+4}×\sqrt{25+196+4}}$ | M1 A1
$= \frac{27}{\sqrt{45×15}} = \frac{9}{3\sqrt{5}×5} = \frac{3}{5\sqrt{5}} = \frac{3}{25}\sqrt{5}$ | M1 A17. Relative to a fixed origin, two lines have the equations\\
and
$$\begin{aligned}
& \mathbf { r } = \left( \begin{array} { c }
7 \\
0 \\
- 3
\end{array} \right) + s \left( \begin{array} { c }
5 \\
4 \\
- 2
\end{array} \right) \\
& \mathbf { r } = \left( \begin{array} { l }
a \\
6 \\
3
\end{array} \right) + t \left( \begin{array} { c }
- 5 \\
14 \\
2
\end{array} \right) ,
\end{aligned}$$
where $a$ is a constant and $s$ and $t$ are scalar parameters.\\
Given that the two lines intersect,\\
(i) find the position vector of their point of intersection,\\
(ii) find the value of $a$.
Given also that $\theta$ is the acute angle between the lines,\\
(iii) find the value of $\cos \theta$ in the form $k \sqrt { 5 }$ where $k$ is rational.\\
\hfill \mbox{\textit{OCR C4 Q7 [10]}}