CAIE P1 2016 June — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeVector operations and magnitudes
DifficultyModerate -0.8 This is a straightforward vector question requiring basic operations: finding AB by subtraction, using AB=BC to find position vector of C, then calculating magnitude and unit vector. All steps are routine procedures with no problem-solving insight needed, making it easier than average but not trivial due to 3D arithmetic.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication
3 Relative to an origin \(O\), the position vectors of points \(A\) and \(B\) are given by $$\overrightarrow { O A } = 2 \mathbf { i } - 5 \mathbf { j } - 2 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 4 \mathbf { i } - 4 \mathbf { j } + 2 \mathbf { k }$$ The point \(C\) is such that \(\overrightarrow { A B } = \overrightarrow { B C }\). Find the unit vector in the direction of \(\overrightarrow { O C }\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\overrightarrow{AB} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}\) or \(\overrightarrow{AC} = 4\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}\)B1
\(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = 6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\)M1 Correct method for \(\overrightarrow{OC}\)
OR \(\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}x-4\\y+4\\z-2\end{pmatrix}\)B1
\(\overrightarrow{OC} = \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}6\\-3\\6\end{pmatrix}\)M1
OR \(\overrightarrow{OC} = 2\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}8\\-8\\4\end{pmatrix} - \begin{pmatrix}2\\-5\\-2\end{pmatrix} = \begin{pmatrix}6\\-3\\6\end{pmatrix}\)B1, M1
Unit vector = (Their \(\overrightarrow{OC}\)) \(\div\) (Mod their \(\overrightarrow{OC}\))M1 Divides by their mod of their \(\overrightarrow{OC}\)
\(= (6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) \div 9\)A1 Correct unsimplified expression 
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ or $\overrightarrow{AC} = 4\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}$ | B1 | |
| $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = 6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ | M1 | Correct method for $\overrightarrow{OC}$ |
| **OR** $\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}x-4\\y+4\\z-2\end{pmatrix}$ | B1 | |
| $\overrightarrow{OC} = \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}6\\-3\\6\end{pmatrix}$ | M1 | |
| **OR** $\overrightarrow{OC} = 2\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}8\\-8\\4\end{pmatrix} - \begin{pmatrix}2\\-5\\-2\end{pmatrix} = \begin{pmatrix}6\\-3\\6\end{pmatrix}$ | B1, M1 | |
| Unit vector = (Their $\overrightarrow{OC}$) $\div$ (Mod their $\overrightarrow{OC}$) | M1 | Divides by their mod of their $\overrightarrow{OC}$ |
| $= (6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) \div 9$ | A1 | Correct unsimplified expression |
3 Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by

$$\overrightarrow { O A } = 2 \mathbf { i } - 5 \mathbf { j } - 2 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 4 \mathbf { i } - 4 \mathbf { j } + 2 \mathbf { k }$$

The point $C$ is such that $\overrightarrow { A B } = \overrightarrow { B C }$. Find the unit vector in the direction of $\overrightarrow { O C }$.

\hfill \mbox{\textit{CAIE P1 2016 Q3 [4]}}
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