Answer
The angle between $ v=\text{5}i+2j\text{ and }w=-3i+j $ is ${{139.8}^{\circ }}$.
Work Step by Step
Consider the vectors, $ v=\text{5}i+2j\text{ and }w=-3i+j $.
The angle between $ v=\text{5}i+2j\text{ and }w=-3i+j $ is,
$\theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left\| v \right\|\times \left\| w \right\|}$
To find the value of $\left\| v \right\|\text{ and }\left\| w \right\|$,
$\begin{align}
& \left\| v \right\|=\sqrt{{{5}^{2}}+{{2}^{2}}} \\
& =\sqrt{25+4} \\
& =\sqrt{29}
\end{align}$
And
$\begin{align}
& \left\| w \right\|=\sqrt{{{\left( -3 \right)}^{2}}+{{1}^{2}}} \\
& =\sqrt{9+1} \\
& =\sqrt{10}
\end{align}$
Substitute the values of $\left\| v \right\|\text{ and }\left\| w \right\|$ in $\theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left\| v \right\|\times \left\| w \right\|}$ and solve for $\theta $,
$\begin{align}
& \theta ={{\cos }^{-1}}\frac{\left( 5i+2j \right)\times \left( -3i+j \right)}{\left( \sqrt{29} \right)\left( \sqrt{10} \right)} \\
& ={{\cos }^{-1}}\frac{-15+2}{\left( \sqrt{29} \right)\left( \sqrt{10} \right)} \\
& ={{\cos }^{-1}}\frac{-13}{\sqrt{290}}\approx {{139.8}^{\circ }}
\end{align}$
Hence, the angle between $ v=\text{5}i+2j\text{ and }w=-3i+j $ is ${{139.8}^{\circ }}$.
