| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Parallel or perpendicular vectors condition |
| Difficulty | Moderate -0.3 Part (i) requires basic trigonometry to find a bearing from vector components (arctan calculation plus angle adjustment). Part (ii) involves recognizing that parallel vectors are scalar multiples, leading to a simple linear equation in k. Both parts are routine applications of standard techniques with no conceptual challenges, making this slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| \(270 - \arctan\left(\frac{6}{4}\right)\) | M1 | Award for \(\arctan p\) seen where \(p = \pm\frac{6}{4}\) or \(\frac{4}{6}\), or equivalent |
| \(= 213.69...\) so \(214°\) | A1 [2] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Need \((-4+3k)\mathbf{i} + (-6-2k)\mathbf{j} = \lambda(7\mathbf{i} - 9\mathbf{j})\) | M1 | Attempt to get LHS in the direction of \((7\mathbf{i} - 9\mathbf{j})\). Could be done by finding (tangents of) angles. Accept the use of \(\lambda = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{-4+3k}{-6-2k} = \frac{7}{-9}\) or equivalent | M1 | Attempt to solve their expression. Allow \(= \frac{7}{9}, \frac{9}{7}, -\frac{9}{7}\) |
| A1 | Expression correct | |
| \(k = 6\) | A1 [4] | Award full marks for \(k = 6\) found WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \(-4 + 3k = 7\lambda\) | M1 | Attempt to solve their equations. Must have both equations |
| \(-6 - 2k = -9\lambda\) | A1 | Correct equations |
| \(k = 6\) | A1 | Award full marks for \(k = 6\) found WWW |
| Answer | Marks |
|---|---|
| M1 | Any attempt to find the value of \(k\) and 'test' |
| M1 | Systematic attempt in (the equivalent of) their expression |
| Full marks for \(k = 6\) found WWW |
## Question 3:
### Part (i)
$270 - \arctan\left(\frac{6}{4}\right)$ | M1 | Award for $\arctan p$ seen where $p = \pm\frac{6}{4}$ or $\frac{4}{6}$, or equivalent
$= 213.69...$ so $214°$ | A1 [2] | cao
### Part (ii)
Need $(-4+3k)\mathbf{i} + (-6-2k)\mathbf{j} = \lambda(7\mathbf{i} - 9\mathbf{j})$ | M1 | Attempt to get LHS in the direction of $(7\mathbf{i} - 9\mathbf{j})$. Could be done by finding (tangents of) angles. Accept the use of $\lambda = 1$
**Either:**
$\frac{-4+3k}{-6-2k} = \frac{7}{-9}$ or equivalent | M1 | Attempt to solve their expression. Allow $= \frac{7}{9}, \frac{9}{7}, -\frac{9}{7}$
| A1 | Expression correct
$k = 6$ | A1 [4] | Award full marks for $k = 6$ found WWW
**Or:**
$-4 + 3k = 7\lambda$ | M1 | Attempt to solve their equations. Must have both equations
$-6 - 2k = -9\lambda$ | A1 | Correct equations
$k = 6$ | A1 | Award full marks for $k = 6$ found WWW
**Trial and error method:**
| M1 | Any attempt to find the value of $k$ and 'test'
| M1 | Systematic attempt in (the equivalent of) their expression
| Full marks for $k = 6$ found WWW
---3 In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are pointing east and north respectively.\\
(i) Calculate the bearing of the vector $- 4 \mathbf { i } - 6 \mathbf { j }$.
The vector $- 4 \mathbf { i } - 6 \mathbf { j } + k ( 3 \mathbf { i } - 2 \mathbf { j } )$ is in the direction $7 \mathbf { i } - 9 \mathbf { j }$.\\
(ii) Find $k$.
\hfill \mbox{\textit{OCR MEI M1 Q3 [6]}}