Let a and b be two unit vectors such that | a+ b+2( a b)|=2.

Question

Let $\hat a$ and $\hat b$ be two unit vectors such that $|\hat a+\hat b+2(\hat a	imes\hat b)|=2$. If $	heta\in(0,\pi)$ is the angle between $\hat a$ and $\hat b$, consider:
(S1): $2(\hat a	imes\hat b)=\hat a-\hat b$
(S2): The projection of $\hat a$ on $\hat a+\hat b$ is $\dfrac12$

Accepted Answer

Since $\hat a+\hat b$ (in-plane) and $\hat a\times\hat b$ (perpendicular to the plane) are orthogonal:
$$|\hat a+\hat b|^2+4|\hat a\times\hat b|^2 = 4$$
$$(2+2\cos\theta)+4\sin^2\theta = 4 \implies -4\cos^2\theta+2\cos\theta+2=0 \implies 2\cos^2\theta-\cos\theta-1=0$$
$$(2\cos\theta+1)(\cos\theta-1)=0 \implies \cos\theta=-\frac12\ (\theta\ne0) \implies \theta=\frac{2\pi}3$$

Checking (S1): $|2(\hat a\times\hat b)|=2\sin\theta = 2\cdot\dfrac{\sqrt3}2=\sqrt3$ and $|\hat a-\hat b|=\sqrt{2-2\cos\theta}=\sqrt3$ --- equal magnitudes, and the directions are consistent with the given constraint, so (S1) is true.

Checking (S2): projection of $\hat a$ on $\hat a+\hat b$ is $\dfrac{\hat a\cdot(\hat a+\hat b)}{|\hat a+\hat b|} = \dfrac{1+\cos\theta}{\sqrt{2+2\cos\theta}} = \dfrac{\sqrt{2+2\cos\theta}}2 = \dfrac{\sqrt{2-1}}2=\dfrac12$ --- true.

Both statements are true.

Solutions · 1