Answer
$a)$ proj$_{\vec{v}}$ $\vec{u}=\langle9,6\rangle$
$b)$ $\vec{u}_{1}=\langle9,6\rangle$ $;$ $\vec{u}_{2}=\langle2,-3\rangle$
Work Step by Step
$\vec{u}=\langle11,3\rangle$ $,$ $\vec{v}=\langle-3,-2\rangle$
$a)$
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:
$\vec{u}\cdot\vec{v}=(11)(-3)+(3)(-2)=-33-6=-39$
Find $|\vec{v}|^{2}$:
$|\vec{v}|^{2}=\Big(\sqrt{(-3)^{2}+(-2)^{2}}\Big)^{2}=(-3)^{2}+(-2)^{2}=9+4=13$
Substitute the known values into the formula and evaluate:
proj$_{\vec{v}}$ $\vec{u}=\Big(\dfrac{-39}{13}\Big)\langle-3,-2\rangle=(-3)\langle-3,-2\rangle=\langle9,6\rangle$
$b)$
Since $\vec{u}_{1}=$proj$_{\vec{v}}$ $\vec{u}$, then $\vec{u}_{1}=\langle9,6\rangle$
Since $\vec{u}_{2}=\vec{u}-$proj$_{\vec{v}}$ $\vec{u}$, substitute the known values into the formula and evaluate to obtain $\vec{u}_{2}$:
$\vec{u}_{2}=\langle11,3\rangle-\langle9,6\rangle=\langle2,-3\rangle$
