facas9
2021-01-02
Answered

Find the angle between the vectors $<1,3>{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}<-2,4>$

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Jozlyn

Answered 2021-01-03
Author has **85** answers

Remember that the angle between two vectors u and v is defined by $\mathrm{cos}\left(\theta \right)=\frac{u\cdot v}{\left|\left|u\right|\right|\left|\left|v\right|\right|}$

So lets calculate$u\cdot v,\left|\left|u\right|\right|,{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\left|\left|v\right|\right|f{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}u=\u27e81,3\u27e9{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}v=\u27e8-2,4\u27e9$

$u\cdot v=\u27e81,3\u27e9\cdot \u27e8-2,4\u27e9=-2+12=10$

$\left|\left|u\right|\right|=\sqrt{{1}^{2}+{3}^{2}}=\sqrt{10}$

$\left|\left|v\right|\right|=\sqrt{{(-2)}^{2}+{4}^{2}}=\sqrt{20}$

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$\mathrm{cos}\left(\theta \right)=\frac{1}{\sqrt{2}}\text{}at\text{}{45}^{\circ}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{315}^{\circ}$ . Because the angle between two vectors must be between $0}^{\circ}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{180}^{\circ$ , that means that the angle between vectors u and v is $45}^{\circ$

So lets calculate

Now plugging these into our equation, we find that

Looking at the unit circle, we see that

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