$u = (1, 1, 0, 2), v = (-1, 0, 3, 4)$. Determine two perpendicular vectors $a$ and $b$ such that $a$ is parallel to $v$ and $u = a + b$ How can I solve this kind of problem? I tried testing with its cross product and trying to get $(u \times v)\cdot a=0$. But I think that's not correct.
 A: We have
$$a=kv=(-k,0,3k,4k)$$
$$b=u-a=(1+k,1,-3k,2-4k)$$
and $$a \cdot b=(1+k)(-k)+(-3k)(3k)+(2-4k)(4k)=0$$
from which we can find $k$.
A: Note $\mathbf u$ is in the linear span of the orthogonal Hamel basis $\mathcal B:=\{\mathbf v,\mathbf w\}$ where $$\mathbf w = (\mathbf v\cdot\mathbf v)\mathbf u-(\mathbf u\cdot\mathbf v)\mathbf v=\langle26,26,0,52\rangle-\langle-7,0,21,28\rangle=\langle33,26,-21,24\rangle$$
$$\therefore\;\;\;\mathbf u=\frac{\mathbf u\cdot\mathbf v}{\mathbf v\cdot\mathbf v}\mathbf v+\frac{\mathbf u\cdot\mathbf w}{\mathbf w\cdot\mathbf w}\mathbf w$$
The rest is direct calculation where $\mathbf u\cdot\mathbf u=6$, $\mathbf u\cdot\mathbf v=7$, $\mathbf v\cdot\mathbf v=26$, $\mathbf u\cdot\mathbf w=156-49=107$, $\mathbf w\cdot\mathbf w=26^2*6-1274=2782$.
$$\therefore\;\;\;\mathbf u=\frac{7}{26}\langle-1,0,3,4\rangle+\frac{107}{2782}\langle33,26,-21,24\rangle$$ $$\therefore\;\;\;\;\;\mathbf a=\frac{7}{26}\langle-1,0,3,4\rangle\;\;\;\;\mathbf b=\frac{1}{26}\langle33,26,-21,24\rangle$$
