Answer
$a)$ proj$_{\vec{v}}$ $\vec{u}=\Big\langle\dfrac{2}{5},-\dfrac{1}{5}\Big\rangle$
$b)$ $\vec{u}_{1}=\Big\langle\dfrac{2}{5},-\dfrac{1}{5}\Big\rangle$ $;$ $\vec{u}_{2}=\Big\langle\dfrac{3}{5},\dfrac{6}{5}\Big\rangle$
Work Step by Step
$\vec{u}=\langle1,1\rangle$ $,$ $\vec{v}=\langle2,-1\rangle$
$a)$
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:
$\vec{u}\cdot\vec{v}=(1)(2)+(1)(-1)=2-1=1$
Find $|\vec{v}|^{2}$:
$|\vec{v}|^{2}=\Big(\sqrt{2^{2}+(-1)^{2}}\Big)^{2}=2^{2}+(-1)^{2}=4+1=5$
Substitute the known values into the formula and evaluate:
proj$_{\vec{v}}$ $\vec{u}=\Big(\dfrac{1}{5}\Big)\langle2,-1\rangle=\Big\langle\dfrac{2}{5},-\dfrac{1}{5}\Big\rangle$
$b)$
Since $\vec{u}_{1}=$proj$_{\vec{v}}$ $\vec{u}$, then $\vec{u}_{1}=\Big\langle\dfrac{2}{5},-\dfrac{1}{5}\Big\rangle$
Since $\vec{u}_{2}=\vec{u}-$proj$_{\vec{v}}$ $\vec{u}$, substitute the known values into the formula and evaluate to obtain $\vec{u}_{2}$:
$\vec{u}_{2}=\langle1,1\rangle-\Big\langle\dfrac{2}{5},-\dfrac{1}{5}\Big\rangle=\Big\langle1-\dfrac{2}{5},1+\dfrac{1}{5}\Big\rangle=\Big\langle\dfrac{3}{5},\dfrac{6}{5}\Big\rangle$
