WAEC 2013 · Paper 2 · Q5

(a)

1. Use $\cos\theta = \dfrac{\mathbf a \cdot \mathbf b}{|\mathbf a||\mathbf b|}$ with $\mathbf a = 2\mathbf i - \mathbf j$ and $\mathbf b = \mathbf i - \mathbf j$.
2. Dot product: $(2)(1) + (-1)(-1) = 3$.
3. Magnitudes: $|\mathbf a| = \sqrt{4 + 1} = \sqrt5$ and $|\mathbf b| = \sqrt{1 + 1} = \sqrt2$.
4. So $\cos\theta = \dfrac{3}{\sqrt5\sqrt2}$
   $= \dfrac{3}{\sqrt{10}}$
   $\approx 0.9487$.
5. Take the inverse cosine: $\theta \approx 18.43^\circ$.
6. The acute angle between the vectors is $18.43^\circ$.