OCR MEI C4 2006 January — Question 3 6 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2006
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeArea of triangle from given side vectors or coordinates
DifficultyModerate -0.3 This is a straightforward application of the scalar product to verify perpendicularity (checking if BA·BC = 0), followed by a simple area calculation using ½|BA||BC|. The question explicitly tells students what to do ('calculate a suitable scalar product') and requires only routine vector arithmetic with no problem-solving insight needed. Slightly easier than average due to the guided approach, though the 3D context and multi-step nature keep it close to typical difficulty.
Spec1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry4.04c Scalar product: calculate and use for angles
3 A triangle ABC has vertices \(\mathrm { A } ( - 2,4,1 ) , \mathrm { B } ( 2,3,4 )\) and \(\mathrm { C } ( 4,8,3 )\). By calculating a suitable scalar product, show that angle ABC is a right angle. Hence calculate the area of the triangle.
Question 3: Verify inequality for Party E
AnswerMarks Guidance
AnswerMark Guidance
Party E (Party 5): \(V_5 = 11.2\), \(N_5 = 2\) (seats allocated)B1 Values stated or used correctly
\(\frac{V_5}{N_5+1} = \frac{11.2}{3} = 3.7\overline{3}\) and \(\frac{V_5}{N_5} = \frac{11.2}{2} = 5.6\)M1 Both fractions evaluated
Since \(3.7\overline{3} < 11 \leq 5.6\) is false — actually \(a=11\): \(3.73 < 11\) ✓ and \(11 \leq 5.6\) ✗
Correct verification that \(\frac{11.2}{3} < 11 \leq \frac{11.2}{2}\), i.e. \(3.73 < 11\) and \(11 \leq 5.6\) with correct conclusionA1 Both inequalities checked with \(a=11\) 
# Question 3: Verify inequality for Party E

| Answer | Mark | Guidance |
|--------|------|----------|
| Party E (Party 5): $V_5 = 11.2$, $N_5 = 2$ (seats allocated) | B1 | Values stated or used correctly |
| $\frac{V_5}{N_5+1} = \frac{11.2}{3} = 3.7\overline{3}$ and $\frac{V_5}{N_5} = \frac{11.2}{2} = 5.6$ | M1 | Both fractions evaluated |
| Since $3.7\overline{3} < 11 \leq 5.6$ is **false** — actually $a=11$: $3.73 < 11$ ✓ and $11 \leq 5.6$ ✗ | | |
| Correct verification that $\frac{11.2}{3} < 11 \leq \frac{11.2}{2}$, i.e. $3.73 < 11$ and $11 \leq 5.6$ with correct conclusion | A1 | Both inequalities checked with $a=11$ |

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3 A triangle ABC has vertices $\mathrm { A } ( - 2,4,1 ) , \mathrm { B } ( 2,3,4 )$ and $\mathrm { C } ( 4,8,3 )$. By calculating a suitable scalar product, show that angle ABC is a right angle. Hence calculate the area of the triangle.

\hfill \mbox{\textit{OCR MEI C4 2006 Q3 [6]}}
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