$ v = [{1, 1, 0}] $ and $ w = [-1, 0, 1] $. Find such vector $ u \in \langle v, v \times w \rangle $, that:
$$ 1) \space \space \alpha(u, v) = \frac{\pi}{3} \\ 2) \space \space \alpha(u, v \times w) = \frac{\pi}{6} \\ 3) \space \space ||u|| = \sqrt{2} $$
What I've already got:
$$ u = [{a, b, c}] \\ v \times w = [{1, -1, 1}] \\ 1) \space \space \cos{\frac{\pi}{3}} = \frac{1}{2} = \frac{a+b}{\sqrt{a^2+b^2+c^2} * \sqrt{2}} \\ 2) \space \space \cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2} = \frac{a-b+c}{\sqrt{a^2+b^2+c^2} * \sqrt{3}} \\ 3) \space \space ||u|| = \sqrt{a^2+b^2+c^2} = \sqrt{2} $$
And from that I guess is't impossible to evaluate a, b and c. I'm not sure how to interpret $ u \in \langle v, v \times w \rangle $ and maybe that's why I stuck here. Could anyone help me with that one?
The set $\langle v,v\times w\rangle$ is the space of all linear combinations of $v$ and $v\times w(=[1,-1,1])$. So, for some $a,b\in\Bbb R$, $u=av+bv\times w=([a+b,a-b,b])$. And $a$ and $b$ must be such that$$\left\{\begin{array}{l}\frac{[a+b,a-b,b].[1,1,0]}{\|[a+b,a-b,b]\|.\|[1,1,0]\|}=\frac12\\\frac{[a+b,a-b,b].[1,-1,1]}{\|[a+b,a-b,b]\|.\|[1,-1,1]\|}=\frac{\sqrt3}2\\\|[a+b,a-b,b]\|=2.\end{array}\right.$$Can you take it from here?