Vector operations and magnitudes

9 Relative to an origin \(O\), the position vectors of the points \(A , B , C\) and \(D\) are given by $$\overrightarrow { O A } = \left( \begin{array} { r } 1 \\ 3 \\ - 1 \end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r } 3 \\ - 1 \\ 3 \end{array} \right) , \quad \overrightarrow { O C } = \left( \begin{array} { l } 4 \\ 2 \\ p \end{array} \right) \quad \text { and } \quad \overrightarrow { O D } = \left( \begin{array} { r } - 1 \\ 0 \\ q \end{array} \right) ,$$ where \(p\) and \(q\) are constants. Find

1. the unit vector in the direction of \(\overrightarrow { A B }\),
2. the value of \(p\) for which angle \(A O C = 90 ^ { \circ }\),
3. the values of \(q\) for which the length of \(\overrightarrow { A D }\) is 7 units.

6 Relative to an origin \(O\), the position vectors of the points \(A\) and \(B\) are given by $$\overrightarrow { O A } = 2 \mathbf { i } - 8 \mathbf { j } + 4 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 7 \mathbf { i } + 2 \mathbf { j } - \mathbf { k }$$

1. Find the value of \(\overrightarrow { O A } \cdot \overrightarrow { O B }\) and hence state whether angle \(A O B\) is acute, obtuse or a right angle.
2. The point \(X\) is such that \(\overrightarrow { A X } = \frac { 2 } { 5 } \overrightarrow { A B }\). Find the unit vector in the direction of \(O X\).

10
\includegraphics[max width=\textwidth, alt={}, center]{56d4d40a-32f5-4f2d-938e-a24312cd42e7-4_552_629_842_758} The diagram shows the parallelogram \(O A B C\). Given that \(\overrightarrow { O A } = \mathbf { i } + 3 \mathbf { j } + 3 \mathbf { k }\) and \(\overrightarrow { O C } = 3 \mathbf { i } - \mathbf { j } + \mathbf { k }\), find

1. the unit vector in the direction of \(\overrightarrow { O B }\),
2. the acute angle between the diagonals of the parallelogram,
3. the perimeter of the parallelogram, correct to 1 decimal place. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

8 Relative to the origin \(O\), the position vectors of the points \(A , B\) and \(C\) are given by $$\overrightarrow { O A } = \left( \begin{array} { l } 2
3
5 \end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { l } 4
2
3 \end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { r }

2 Relative to an origin \(O\), the position vectors of the points \(A , B\) and \(C\) are given by $$\overrightarrow { O A } = \left( \begin{array} { r } 2
- 1

5 Relative to an origin \(O\), the position vectors of the points \(A , B\) and \(C\) are given by $$\overrightarrow { O A } = \left( \begin{array} { r } 3
2
- 3 \end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r } 5
- 1
- 2 \end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }

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 \(\overrightarrow { O C }\).

10 Relative to an origin \(O\), the position vectors of the points \(A , B\) and \(X\) are given by $$\overrightarrow { O A } = \left( \begin{array} { r } - 8
- 4
2 \end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r } 10
2

6 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }