Answer
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{4}{6}(-1,2,-1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
= \frac{4}{6}( 1,2,-1).$$
Work Step by Step
Let $u=(1,2,-1)$, $v=(-1,2,-1)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have
$$\langle{u}, {v}\rangle=4, \quad \langle{u}, {u}\rangle=6, \quad \langle{v}, {v}\rangle=6.$$
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{4}{6}(-1,2,-1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{4}{6}( 1,2,-1).$$