Answer
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
= \frac{-5}{10}(-1,1,2,2).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{-5}{46}(0,1,3,-6).$$
Work Step by Step
Let $u=(0,1,3,-6)$, $v=(-1,1,2,2)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have
$$\langle{u}, {v}\rangle=-5, \quad \langle{u}, {u}\rangle=46, \quad \langle{v}, {v}\rangle=10.$$
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{-5}{10}(-1,1,2,2).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{-5}{46}(0,1,3,-6).$$
