Answer
(a) $-\sqrt 2$
(b) $\langle -1, -1 \rangle$
(c) $\langle -1, -1 \rangle$, $\langle -7, 7 \rangle$
Work Step by Step
(a) Given $\vec u=\langle -8, 6 \rangle$ and $\vec v=\langle 20, 20 \rangle$，we have
$comp_v\vec u=\frac{\vec u\cdot\vec v}{|\vec v|}=\frac{-8\times20+6\times20}{\sqrt {20^2+20^2}}=-\frac{40}{20\sqrt 2}=-\sqrt 2$
(b) $proj_v\vec u=(\frac{\vec u\cdot\vec v}{|\vec v|^2})\vec v=-\frac{40}{800}\langle 20, 20 \rangle=\langle -1, -1 \rangle$
(c) Clearly $\vec u_1=proj_v\vec u=\langle -1, -1 \rangle$, and $\vec u_2=\vec u-proj_v\vec u=\langle -8+1, 6+1 \rangle=\langle -7, 7 \rangle$
