| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Moderate -0.3 This is a straightforward three-part vector question testing standard techniques: finding a unit vector (routine division by magnitude), using parallelism conditions (setting i and j components to zero), and applying the scalar product formula for angles. All parts are textbook exercises requiring direct application of formulas with no problem-solving insight, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | \mathbf{a} | = \sqrt{2^2 + 6^2 + 9^2} = 11\) |
| A unit vector is \(\pm\frac{1}{11}\begin{pmatrix}2\\6\\9\end{pmatrix}\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2\\6\\9\end{pmatrix} + \lambda\begin{pmatrix}1\\3\\-1\end{pmatrix} = k\begin{pmatrix}0\\0\\1\end{pmatrix}\) | M1 | |
| \(\Rightarrow \begin{cases}0 = 2+\lambda \\ 0 = 6+3\lambda \\ (k = 9-\lambda)\end{cases}\) | A1 | |
| \(\Rightarrow \lambda = -2\), since that satisfies both \(x\) and \(y\) component relations | ||
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Angle is \(\theta\) where \(\theta = \cos^{-1}\frac{ | \mathbf{a}\cdot\mathbf{b} | }{ |
| \(= \cos^{-1}\frac{2+18-9}{11\times\sqrt{1^2+3^2+1^2}} = \cos^{-1}\frac{1}{\sqrt{11}}\) | B1, B1 | scalar product; B1 for \( |
| \(= 72.45...\) | ||
| \(= 72.5°\) to 1 dp | B1 | |
| Total: 3 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|\mathbf{a}| = \sqrt{2^2 + 6^2 + 9^2} = 11$ | M1 | |
| A unit vector is $\pm\frac{1}{11}\begin{pmatrix}2\\6\\9\end{pmatrix}$ | A1 | |
| **Total: 2** | | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\6\\9\end{pmatrix} + \lambda\begin{pmatrix}1\\3\\-1\end{pmatrix} = k\begin{pmatrix}0\\0\\1\end{pmatrix}$ | M1 | |
| $\Rightarrow \begin{cases}0 = 2+\lambda \\ 0 = 6+3\lambda \\ (k = 9-\lambda)\end{cases}$ | A1 | |
| $\Rightarrow \lambda = -2$, since that satisfies both $x$ and $y$ component relations | | |
| **Total: 2** | | |
---
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Angle is $\theta$ where $\theta = \cos^{-1}\frac{|\mathbf{a}\cdot\mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}$ | | |
| $= \cos^{-1}\frac{2+18-9}{11\times\sqrt{1^2+3^2+1^2}} = \cos^{-1}\frac{1}{\sqrt{11}}$ | B1, B1 | scalar product; B1 for $|\mathbf{b}|$ |
| $= 72.45...$| | |
| $= 72.5°$ to 1 dp | B1 | |
| **Total: 3** | | |
---4 You are given that $\mathbf { a } = 2 \mathbf { i } + 6 \mathbf { j } + 9 \mathbf { k }$ and $\mathbf { b } = \mathbf { i } + 3 \mathbf { j } - \mathbf { k }$.\\
(i) Write down a unit vector parallel to a.\\
(ii) Find the value of $\lambda$ such that $\mathbf { a } + \lambda \mathbf { b }$ is parallel to $\mathbf { k }$.\\
(iii) Calculate the size of the angle between $\mathbf { a }$ and $\mathbf { b }$.
\hfill \mbox{\textit{OCR MEI C4 Q4 [7]}}