Answer
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
= \frac{-19}{22}(4,-1,2,-1)$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
= \frac{-19}{40}(-1,4,-2,3).$$
Work Step by Step
Let $u=(-1,4,-2,3)$, $v=(4,-1,2,-1)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have
$$\langle{u}, {v}\rangle=-19, \quad \langle{u}, {u}\rangle=40, \quad \langle{v}, {v}\rangle=22.$$
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{-19}{22}(4,-1,2,-1)$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{-19}{40}(-1,4,-2,3).$$