Answer
$a)$ proj$_{\vec{v}}$ $\vec{u}=\Big\langle-\dfrac{1}{2},\dfrac{3}{2}\Big\rangle$
$b)$ $\vec{u}_{1}=\Big\langle-\dfrac{1}{2},\dfrac{3}{2}\Big\rangle$ $;$ $\vec{u}_{2}=\Big\langle\dfrac{3}{2},\dfrac{1}{2}\Big\rangle$
Work Step by Step
$\vec{u}=\langle1,2\rangle$ $,$ $\vec{v}=\langle1,-3\rangle$
$a)$
The projection of $\vec{u}$ onto $\vec{v}$ is given by proj$_{\vec{v}}$ $\vec{u}=\Big(\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|^{2}}\Big)\vec{v}$.
Find $\vec{u}\cdot\vec{v}$ by multiplying corresponding components and adding:
$\vec{u}\cdot\vec{v}=(1)(1)+(2)(-3)=1-6=-5$
Find $|\vec{v}|^{2}$:
$|\vec{v}|^{2}=\Big(\sqrt{1^{2}+(-3)^{2}}\Big)^{2}=1^{2}+(-3)^{2}=1+9=10$
Substitute the known values into the formula and evaluate:
proj$_{\vec{v}}$ $\vec{u}=\Big(\dfrac{-5}{10}\Big)\langle1,-3\rangle=-\Big(\dfrac{1}{2}\Big)\langle1,-3\rangle=\Big\langle-\dfrac{1}{2},\dfrac{3}{2}\Big\rangle$
$b)$
Since $\vec{u}_{1}=$proj$_{\vec{v}}$ $\vec{u}$, then $\vec{u}_{1}=\Big\langle-\dfrac{1}{2},\dfrac{3}{2}\Big\rangle$
Since $\vec{u}_{2}=\vec{u}-$proj$_{\vec{v}}$ $\vec{u}$, substitute the known values into the formula and evaluate to obtain $\vec{u}_{2}$:
$\vec{u}_{2}=\langle1,2\rangle-\Big\langle-\dfrac{1}{2},\dfrac{3}{2}\Big\rangle=\Big\langle1+\dfrac{1}{2},2-\dfrac{3}{2}\Big\rangle=\Big\langle\dfrac{3}{2},\dfrac{1}{2}\Big\rangle$
