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Let $u$, $v$ be unit vectors in different directions and we denote the angle between $u$ and $v$ by $\theta$ and the angle between $u$ and $u + v$ by $\varphi$.

(a) Show that $\vert u + v \vert = 2 \cos \theta $

(b) Show that $u + v$ bisects the angle between $u$ and $v$ internally, that is, $\varphi = \dfrac{\theta}{2}$

I've tried squaring $ {\vert u+v \vert }^2=(u+v) \cdot (u+v)$ ... not sure what to do.

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    You're on the right track: Now expand $({\bf u} + {\bf v}) \cdot ({\bf u} + {\bf v})$ using linearity and then use the fact that ${\bf u}, {\bf v}$ are unit vectors$. – 2017-02-15

1 Answers 1

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By Law of cosines

$$ ||v||^2 = ||u||^2 + ||u+v||^2 - 2 ||u|| ||u+v|| \cos \varphi \implies ||u+v|| = 2 \cos \varphi$$

and

$$ ||u||^2 = ||v||^2 +||u+v||^2 - 2 ||v|| ||u+v|| \cos ( \theta - \varphi ) \implies ||u+v|| = 2 \cos (\theta - \varphi) $$

Thus,

$$ \cos \varphi = \cos ( \theta - \varphi) \implies \varphi = \frac{ \theta }{2} $$