Answer
(a) $\frac{17\sqrt {37}}{37}$
(b) $\langle \frac{102}{37}, -\frac{17}{37} \rangle$
(c) $\langle \frac{102}{37}, -\frac{17}{37} \rangle$, $\langle \frac{9}{37}, \frac{54}{37} \rangle$
Work Step by Step
(a) Given $\vec u=\langle 3, 1 \rangle$ and $vec v=\langle 6, -1 \rangle$，we have
$comp_v\vec u=\frac{\vec u\cdot\vec v}{|\vec v|}=\frac{3\times6-1\times1}{\sqrt {6^2+1^2}}=\frac{17\sqrt {37}}{37}$
(b) $proj_v\vec u=(\frac{\vec u\cdot\vec v}{|\vec v|^2})\vec v=\frac{17}{37}\langle 6, -1 \rangle=\langle \frac{102}{37}, -\frac{17}{37} \rangle$
(c) Clearly $\vec u_1=proj_v\vec u=\langle \frac{102}{37}, -\frac{17}{37} \rangle$, and $\vec u_2=\vec u-proj_v\vec u=\langle 3-\frac{102}{37}, 1+\frac{17}{37} \rangle=\langle \frac{9}{37}, \frac{54}{37} \rangle$
