| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward two-part question on vector lines requiring standard techniques: (i) substituting a point into both equations to find parameters then α, and (ii) using the scalar product formula with cos(60°). Both parts are routine applications of core C4 methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mu = 3\) soi | B1 | From \(3 + 2\lambda = 4 + \mu\) and \(5 + \lambda = 10 - \mu\); NB \(\lambda = 2\) |
| \(1 = 19 + (\text{their }3)\times\alpha\) oe | M1 | Do not allow sign errors |
| \([\alpha =]{-6}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\times1 + 1\times(-1) + 1\times\alpha\) | M1* | Allow 1 sign error; NB \(1+\alpha = \sqrt{6}\times\sqrt{(2+\alpha^2)}\times\cos 60°\) |
| \(\sqrt{(2^2+1^2+1^2)}\times\sqrt{(1^2+(-1)^2+\alpha^2)}\times\cos 60°\) | M1* | Allow 1 slip, e.g. sign error or omission of power |
| e.g. their \(4 + 8\alpha + 4\alpha^2 = 6(2+\alpha^2)\) | M1dep* | Square both sides |
| \(\alpha = 2\) cao | A1 [4] | If M1M1M0, B2 for unsupported or alternative valid method; NB \(2\alpha^2 - 8\alpha + 8 = 0\) |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mu = 3$ soi | B1 | From $3 + 2\lambda = 4 + \mu$ and $5 + \lambda = 10 - \mu$; NB $\lambda = 2$ |
| $1 = 19 + (\text{their }3)\times\alpha$ oe | M1 | Do not allow sign errors |
| $[\alpha =]{-6}$ | A1 **[3]** | |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\times1 + 1\times(-1) + 1\times\alpha$ | M1* | Allow 1 sign error; NB $1+\alpha = \sqrt{6}\times\sqrt{(2+\alpha^2)}\times\cos 60°$ |
| $\sqrt{(2^2+1^2+1^2)}\times\sqrt{(1^2+(-1)^2+\alpha^2)}\times\cos 60°$ | M1* | Allow 1 slip, e.g. sign error or omission of power |
| e.g. their $4 + 8\alpha + 4\alpha^2 = 6(2+\alpha^2)$ | M1dep* | Square both sides |
| $\alpha = 2$ cao | A1 **[4]** | If M1M1M0, B2 for unsupported or alternative valid method; NB $2\alpha^2 - 8\alpha + 8 = 0$ |
---9 Two lines have equations
$$\mathbf { r } = 3 \mathbf { i } + 5 \mathbf { j } - \mathbf { k } + \lambda ( 2 \mathbf { i } + \mathbf { j } + \mathbf { k } ) \text { and } \mathbf { r } = 4 \mathbf { i } + 10 \mathbf { j } + 19 \mathbf { k } + \mu ( \mathbf { i } - \mathbf { j } + \alpha \mathbf { k } ) ,$$
where $\alpha$ is a constant.\\
Find the value of $\alpha$ in each of the following cases.\\
(i) The lines intersect at the point (7,7,1).\\
(ii) The angle between their directions is $60 ^ { \circ }$.
\hfill \mbox{\textit{OCR C4 2015 Q9 [7]}}