Answer
$a)$ proj$_{\vec{v}}$ $\vec{u}=\Big\langle-\dfrac{18}{5},\dfrac{24}{5}\Big\rangle$
$b)$ $\vec{u}_{1}=\Big\langle-\dfrac{18}{5},\dfrac{24}{5}\Big\rangle$ $;$ $\vec{u}_{2}=\Big\langle\dfrac{28}{5},\dfrac{21}{5}\Big\rangle$
Work Step by Step
$\vec{u}=\langle2,9\rangle$ $,$ $\vec{v}=\langle-3,4\rangle$
$a)$
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:
$\vec{u}\cdot\vec{v}=(2)(-3)+(9)(4)=-6+36=30$
Find $|\vec{v}|^{2}$:
$|\vec{v}|^{2}=\Big(\sqrt{(-3)^{2}+4^{2}}\Big)^{2}=(-3)^{2}+4^{2}=9+16=25$
Substitute the known values into the formula and evaluate:
proj$_{\vec{v}}$ $\vec{u}=\Big(\dfrac{30}{25}\Big)\langle-3,4\rangle=\Big(\dfrac{6}{5}\Big)\langle-3,4\rangle=\Big\langle-\dfrac{18}{5},\dfrac{24}{5}\Big\rangle$
$b)$
Since $\vec{u}_{1}=$proj$_{\vec{v}}$ $\vec{u}$, then $\vec{u}_{1}=\Big\langle-\dfrac{18}{5},\dfrac{24}{5}\Big\rangle$
Since $\vec{u}_{2}=\vec{u}-$proj$_{\vec{v}}$ $\vec{u}$, substitute the known values into the formula and evaluate to obtain $\vec{u}_{2}$:
$\vec{u}_{2}=\langle2,9\rangle-\Big\langle-\dfrac{18}{5},\dfrac{24}{5}\Big\rangle=\Big\langle2+\dfrac{18}{5},9-\dfrac{24}{5}\Big\rangle=\Big\langle\dfrac{28}{5},\dfrac{21}{5}\Big\rangle$
