Answer
$a)$ proj$_\vec{v}$ $\vec{u}=\langle4,2\rangle$
$b)$ $\vec{u}_{1}=\langle4,2\rangle$ $;$ $\vec{u}_{2}=\langle3,-6\rangle$
Work Step by Step
$\vec{u}=\langle7,-4\rangle$ $,$ $\vec{v}=\langle2,1\rangle$
$a)$
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:
$\vec{u}\cdot\vec{v}=(7)(2)+(-4)(1)=14-4=10$
Find $|\vec{v}|^{2}$:
$|\vec{v}|^{2}=(\sqrt{2^{2}+1^{2}})^{2}=2^{2}+1^{2}=4+1=5$
Substitute the known values into the formula and evaluate:
proj$_\vec{v}$ $\vec{u}=\Big(\dfrac{10}{5}\Big)\langle2,1\rangle=2\langle2,1\rangle=\langle4,2\rangle$
$b)$
Since $\vec{u}_{1}=$proj$_{\vec{v}}$ $\vec{u}$, then $\vec{u}_{1}=\langle4,2\rangle$
Since $\vec{u}_{2}=\vec{u}-$proj$_{\vec{v}}$ $\vec{u}$, substitute the known values into the formula and evaluate to obtain $\vec{u}_{2}$:
$\vec{u}_{2}=\langle7,-4\rangle-\langle4,2\rangle=\langle7-4,-4-2\rangle=\langle3,-6\rangle$
