CAIE P1 2013 June — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypePerpendicularity conditions
DifficultyModerate -0.3 This is a straightforward multi-part vectors question testing standard techniques: parallel vectors (scalar multiples), perpendicularity (dot product = 0), and unit vectors. All parts are routine applications of basic vector concepts with minimal problem-solving required, making it slightly easier than average but not trivial due to the algebraic manipulation needed in part (ii).
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors

6 Relative to an origin \(O\), the position vectors of points \(A\) and \(B\) are given by $$\overrightarrow { O A } = \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 3 \mathbf { i } + p \mathbf { j } + q \mathbf { k }$$ where \(p\) and \(q\) are constants.
  1. State the values of \(p\) and \(q\) for which \(\overrightarrow { O A }\) is parallel to \(\overrightarrow { O B }\).
  2. In the case where \(q = 2 p\), find the value of \(p\) for which angle \(B O A\) is \(90 ^ { \circ }\).
  3. In the case where \(p = 1\) and \(q = 8\), find the unit vector in the direction of \(\overrightarrow { A B }\).

\(\overrightarrow{OA} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\), \(\overrightarrow{OB} = 3\mathbf{i} + p\mathbf{j} + q\mathbf{k}\)
AnswerMarks Guidance
(i) \(p = -6, q = 6\)B1 B1 [2]
(ii) dot product \(= 0 \rightarrow 3 - 2p + 4q = 0\) → \(p = -1.5\)M1 A1 Use of \(x_1x_2 + y_1y_2 + z_1z_2 = 0\)
(iii) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\). Unit vector \(= (2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) \div 7\)B1 M1 A1 not for \(\mathbf{b} - \mathbf{a}\). M1 for division by modulus. \(\checkmark\) on B1
[2][3]
$\overrightarrow{OA} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$, $\overrightarrow{OB} = 3\mathbf{i} + p\mathbf{j} + q\mathbf{k}$

(i) $p = -6, q = 6$ | B1 B1 | [2] |

(ii) dot product $= 0 \rightarrow 3 - 2p + 4q = 0$ → $p = -1.5$ | M1 A1 | Use of $x_1x_2 + y_1y_2 + z_1z_2 = 0$

(iii) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}$. Unit vector $= (2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) \div 7$ | B1 M1 A1 | not for $\mathbf{b} - \mathbf{a}$. M1 for division by modulus. $\checkmark$ on B1

[2] | [3] |

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6 Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by

$$\overrightarrow { O A } = \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 3 \mathbf { i } + p \mathbf { j } + q \mathbf { k }$$

where $p$ and $q$ are constants.\\
(i) State the values of $p$ and $q$ for which $\overrightarrow { O A }$ is parallel to $\overrightarrow { O B }$.\\
(ii) In the case where $q = 2 p$, find the value of $p$ for which angle $B O A$ is $90 ^ { \circ }$.\\
(iii) In the case where $p = 1$ and $q = 8$, find the unit vector in the direction of $\overrightarrow { A B }$.

\hfill \mbox{\textit{CAIE P1 2013 Q6 [7]}}