OCR C4 2012 June — Question 5 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeAngle between two vectors
DifficultyModerate -0.8 This is a straightforward application of the dot product formula for angles and the magnitude formula. Part (i) requires direct substitution into cos θ = a·b/(|a||b|), and part (ii) uses |a-b|² = |a|² + |b|² - 2a·b. Both are standard textbook exercises with all values given, requiring only routine recall and calculation.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication

5 \includegraphics[max width=\textwidth, alt={}, center]{b5d85e48-0d5a-4edf-bf58-eba4f8d28d3d-2_425_680_1302_689} In the diagram the points \(A\) and \(B\) have position vectors \(\mathbf { a }\) and \(\mathbf { b }\) with respect to the origin \(O\). Given that \(| \mathbf { a } | = 3 , | \mathbf { b } | = 4\) and \(\mathbf { a . b } = 6\), find
  1. the angle \(A O B\),
  2. \(| \mathbf { a } - \mathbf { b } |\).

Question 5:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
Use \(\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{\mathbf{a}
Obtain \(\left(\cos\theta = \frac{6}{12}\right)\), \(\theta = 60°\) or \(\frac{1}{3}\pi\) or 1.05 or betterA1 Better: 1.0471976 (rot)
Total: [2]
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
Indicate \(\mathbf{a}-\mathbf{b}\) is vector joining ends of \(\mathbf{a}\) and \(\mathbf{b}\) or equiv; \(\mathbf{a}-\mathbf{b} =
Use cosine rule correctly on 3, 4 and included angle from (i)M1 Or any other correct method
Obtain \(\sqrt{13}\) or 3.61 or better (No ft from wrong \(\theta\))A1 3.6055513 (rot)
Total: [3]
# Question 5:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ | M1 | |
| Obtain $\left(\cos\theta = \frac{6}{12}\right)$, $\theta = 60°$ or $\frac{1}{3}\pi$ or 1.05 or better | A1 | Better: 1.0471976 (rot) |
| **Total: [2]** | | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Indicate $\mathbf{a}-\mathbf{b}$ is vector joining ends of $\mathbf{a}$ and $\mathbf{b}$ or equiv; $|\mathbf{a}-\mathbf{b}|=|\mathbf{a}|-|\mathbf{b}|$ or anything similar $\to$ M0 | M1 | |
| Use cosine rule correctly on 3, 4 and included angle from (i) | M1 | Or any other correct method |
| Obtain $\sqrt{13}$ or 3.61 or better (No ft from wrong $\theta$) | A1 | 3.6055513 (rot) |
| **Total: [3]** | | |

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5\\
\includegraphics[max width=\textwidth, alt={}, center]{b5d85e48-0d5a-4edf-bf58-eba4f8d28d3d-2_425_680_1302_689}

In the diagram the points $A$ and $B$ have position vectors $\mathbf { a }$ and $\mathbf { b }$ with respect to the origin $O$. Given that $| \mathbf { a } | = 3 , | \mathbf { b } | = 4$ and $\mathbf { a . b } = 6$, find\\
(i) the angle $A O B$,\\
(ii) $| \mathbf { a } - \mathbf { b } |$.

\hfill \mbox{\textit{OCR C4 2012 Q5 [5]}}