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