CAIE P1 2010 November — Question 5 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2010
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeAngles between vectors
DifficultyStandard +0.3 This is a straightforward 3D vectors question requiring students to set up position vectors, find two vectors meeting at C, then apply the scalar product formula for the angle between vectors. While it involves 3D geometry, the setup is clear, the calculations are routine, and it's a standard textbook application making it slightly easier than average.
Spec1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum

5 \includegraphics[max width=\textwidth, alt={}, center]{73c0c113-8f35-4e7f-ad5d-604602498b71-2_741_533_1279_808} The diagram shows a pyramid \(O A B C\) with a horizontal base \(O A B\) where \(O A = 6 \mathrm {~cm} , O B = 8 \mathrm {~cm}\) and angle \(A O B = 90 ^ { \circ }\). The point \(C\) is vertically above \(O\) and \(O C = 10 \mathrm {~cm}\). Unit vectors \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\) are parallel to \(O A , O B\) and \(O C\) as shown. Use a scalar product to find angle \(A C B\).

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\overrightarrow{AC} = -6\mathbf{i} + 10\mathbf{k}\)\(B1\) co (or \(\overrightarrow{CA}\))
\(\overrightarrow{BC} = -8\mathbf{j} + 10\mathbf{k}\)\(B1\) co (or \(\overrightarrow{CB}\))
\(\overrightarrow{AC}.\overrightarrow{BC} = 100\)\(M1\) Must be scalar – available for any pair
\(\overrightarrow{AC}.\overrightarrow{BC} = \sqrt{136}\sqrt{164}\cos ACB\)\(M1\) For modulus – available for any vector
\(M1\)All linked correctly – for \(ACB\) only
Angle \(ACB = 48.0°\)\(A1\) co
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AC} = -6\mathbf{i} + 10\mathbf{k}$ | $B1$ | co (or $\overrightarrow{CA}$) |
| $\overrightarrow{BC} = -8\mathbf{j} + 10\mathbf{k}$ | $B1$ | co (or $\overrightarrow{CB}$) |
| $\overrightarrow{AC}.\overrightarrow{BC} = 100$ | $M1$ | Must be scalar – available for any pair |
| $\overrightarrow{AC}.\overrightarrow{BC} = \sqrt{136}\sqrt{164}\cos ACB$ | $M1$ | For modulus – available for any vector |
| | $M1$ | All linked correctly – for $ACB$ only |
| Angle $ACB = 48.0°$ | $A1$ | co |

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5\\
\includegraphics[max width=\textwidth, alt={}, center]{73c0c113-8f35-4e7f-ad5d-604602498b71-2_741_533_1279_808}

The diagram shows a pyramid $O A B C$ with a horizontal base $O A B$ where $O A = 6 \mathrm {~cm} , O B = 8 \mathrm {~cm}$ and angle $A O B = 90 ^ { \circ }$. The point $C$ is vertically above $O$ and $O C = 10 \mathrm {~cm}$. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O B$ and $O C$ as shown.

Use a scalar product to find angle $A C B$.

\hfill \mbox{\textit{CAIE P1 2010 Q5 [6]}}