Find the angle between the vectors $<1,3>\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}<-2,4>$
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Jozlyn
Remember that the angle between two vectors u and v is defined by $\mathrm{cos}\left(\theta \right)=\frac{u\cdot v}{||u||||v||}$
So lets calculate $u\cdot v,||u||,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}||v||f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}u=⟨1,3⟩\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}v=⟨-2,4⟩$
$u\cdot v=⟨1,3⟩\cdot ⟨-2,4⟩=-2+12=10$
$||u||=\sqrt{{1}^{2}+{3}^{2}}=\sqrt{10}$
$||v||=\sqrt{{\left(-2\right)}^{2}+{4}^{2}}=\sqrt{20}$
Now plugging these into our equation, we find that
$\mathrm{cos}\left(\theta \right)=\frac{10}{\sqrt{10}\sqrt{20}}=\frac{1}{\sqrt{2}}$
Looking at the unit circle, we see that . Because the angle between two vectors must be between ${0}^{\circ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{180}^{\circ }$, that means that the angle between vectors u and v is ${45}^{\circ }$