Answer
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{4}{5}(2,1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{4}{5}( 1,2).$$
Work Step by Step
Let $u=(1,2)$, $v=(2,1)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have
$$\langle{u}, {v}\rangle=4, \quad \langle{u}, {u}\rangle=5, \quad \langle{v}, {v}\rangle=5.$$
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{4}{5}(2,1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{4}{5}( 1,2).$$
(c)