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.