| Date | November 2013 | Marks available | 3 | Reference code | 13N.2.hl.TZ0.7 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The vectors a and b are such that a \( = (3\cos \theta + 6)\)i \( + 7\) j and b \( = (\cos \theta - 2)\)i \( + (1 + \sin \theta )\)j.
Given that a and b are perpendicular,
show that \(3{\sin ^2}\theta - 7\sin \theta + 2 = 0\);
find the smallest possible positive value of \(\theta \).
Markscheme
attempting to form \((3\cos \theta + 6)(\cos \theta - 2) + 7(1 + \sin \theta ) = 0\) M1
\(3{\cos ^2}\theta - 12 + 7\sin \theta + 7 = 0\) A1
\(3\left( {1 - {{\sin }^2}\theta } \right) + 7\sin \theta - 5 = 0\) M1
\(3{\sin ^2}\theta - 7\sin \theta + 2 = 0\) AG
[3 marks]
attempting to solve algebraically (including substitution) or graphically for \(\sin \theta \) (M1)
\(\sin \theta = \frac{1}{3}\) (A1)
\(\theta = 0.340{\text{ }}( = 19.5^\circ )\) A1
[3 marks]
Examiners report
Part (a) was very well done. Most candidates were able to use the scalar product and \({\cos ^2}\theta = 1 - {\sin ^2}\theta \) to show the required result.
Part (b) was reasonably well done. A few candidates confused ‘smallest possible positive value’ with a minimum function value. Some candidates gave \(\theta = 0.34\) as their final answer.
