Answer
(a) $45^{\circ}$
(b) $\frac{\sqrt {26}}{2}$
(c) $\frac{5}{2}i-\frac{1}{2}j$
Work Step by Step
We are given $\vec u=3i+2j$ and $\vec v=5i-j$
(a) The angle between the two vectors can be found as $\cos\theta=\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}
=\frac{3\times5+2(-1)}{\sqrt {(3)^2+(2)^2}\sqrt {(5)^2+(-1)^2 }}=\frac{13}{\sqrt {13}\sqrt {26}}=\frac{\sqrt 2}{2}$ which gives $\theta=45^{\circ}$
(b) The component $comp_v\vec u=\frac{\vec u\cdot\vec v}{|\vec v|}==\frac{3\times5+2(-1)}{\sqrt {5^2+(-1)^2}}=\frac{13}{\sqrt {26}}=\frac{\sqrt {26}}{2}$
(c) $proj_v\vec u=\frac{\vec u\cdot\vec v}{|\vec v|^2}\vec v=\frac{3\times5+2(-1)}{5^2+(-1)^2}(5i-j)=\frac{5}{2}i-\frac{1}{2}j$
