Let $\alpha$ be the angle between the unit vectors a and b, where $0 \leqslant \alpha \leqslant \pi$.

(a)     Express $|$a − b$|$ and $|$a + b$|$ in terms of $\alpha$.

(b)     Hence determine the value of $\cos \alpha$ for which $|$a + b$|$ = 3 $|$a − b$|$.

METHOD 1

(a)     $|$a – b$|$ = $\sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} - 2\left| a \right|\left| b \right|\cos \alpha }$     M1

$= \sqrt {2 - 2\cos \alpha }$     A1

$|$a + b$|$ = $\sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} - 2\left| a \right|\left| b \right|\cos (\pi - \alpha )}$

$= \sqrt {2 + 2\cos \alpha }$     A1

Note: Accept the use of a, b for $|$a$|$, $|$b$|$.

(b)     $= \sqrt {2 + 2\cos \alpha } = 3\sqrt {2 - 2\cos \alpha }$     M1

$\cos \alpha = \frac{4}{5}$     A1 

METHOD 2

(a)     $|$a – b$|$ = $2\sin \frac{\alpha}{2}$     M1A1

$|$a + b$|$ = $2\sin \left( {\frac{\pi }{2} - \frac{\alpha}{2}} \right) = 2\cos \frac{\alpha}{2}$     A1

Note: Accept the use of a, b for $|$a $|$, $|$b$|$.

(b)     $2\cos \frac{\alpha}{2} = 6\sin \frac{\alpha }{2}$

$\tan \frac{\alpha }{2} = \frac{1}{3} \Rightarrow {\cos ^2}\frac{\alpha }{2} = \frac{9}{{10}}$     M1

$\cos \alpha = 2{\cos ^2}\frac{\alpha }{2} - 1 = \frac{4}{5}$     A1

[5 marks]

To solve this problem, candidates had to know either that (a + b)(a + b) = $|$a + b${|^2}$ or that the diagonals of a parallelogram whose sides are a and b represent the vectors a + b and a – b. It was clear from the scripts that many candidates were unaware of either result and were therefore unable to make any progress in this question.