AS June 2018 Paper 1 Q2
2 Find, to the nearest degree, the angle between the vectors \(\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} -2 \\ 3 \\ -3 \end{pmatrix}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\cos\theta = \dfrac{1 \times (-2) + 0 \times 3 + (-2) \times (-3)}{\sqrt{5}\sqrt{22}} = 0.381\ldots\) | M1 | 1.1a |
| A1 | 1.1 | |
| \(\theta = 68^\circ\) to nearest degree | A1 | 1.1 |
| [3] |
Notes
M1: Use of \(\cos\theta = \dfrac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
A1: (1st) \(0.381\ldots\) or \(\dfrac{4}{\sqrt{5}\sqrt{22}}\)
A1: (2nd) must be nearest degree
allow unsupported correct answers