Answer
comp$_{\vec{v}}$ $\vec{u}=\sqrt{2}$
Work Step by Step
$\vec{u}=\langle-3,5\rangle$ $,$ $\vec{v}=\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle$
The component of $\vec{u}$ along $\vec{v}$ is comp$_{\vec{v}}$ $\vec{u}=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|}$
where $\vec{u}\cdot\vec{v}$ is the dot product of the vectors $\vec{u}$ and $\vec{v}$ and $|\vec{v}|$ is the magnitude of the vector $\vec{v}$.
Find $\vec{u}\cdot\vec{v}$ by multiplying corresponding components and adding:
$\vec{u}\cdot\vec{v}=(-3)\Big(\dfrac{1}{\sqrt{2}}\Big)+(5)\Big(\dfrac{1}{\sqrt{2}}\Big)=-\dfrac{3}{\sqrt{2}}+\dfrac{5}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$
Find $|\vec{v}|$:
$|\vec{v}|=\sqrt{\Big(\dfrac{1}{\sqrt{2}}\Big)^{2}+\Big(\dfrac{1}{\sqrt{2}}\Big)^{2}}=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}=\sqrt{\dfrac{2}{2}}=\sqrt{1}=1$
Substitute $\vec{u}\cdot\vec{v}$ and $|\vec{v}|$ into the formula and evaluate:
comp$_{\vec{v}}$ $\vec{u}=\dfrac{\sqrt{2}}{1}=\sqrt{2}$
