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Question Vectors and spaces
ANSWERED Find the angle between the vectors $$\displaystyle<{1},{3}>{\quad\text{and}\quad}<-{2},{4}>$$ 2021-01-03
Remember that the angle between two vectors u and v is defined by $$\displaystyle \cos{{\left(\theta\right)}}=\frac{{{u}\cdot{v}}}{{{\left|{\left|{u}\right|}\right|}{\left|{\left|{v}\right|}\right|}}}$$
So lets calculate $$\displaystyle{u}\cdot{v},{\left|{\left|{u}\right|}\right|},{\quad\text{and}\quad}{\left|{\left|{v}\right|}\right|}{f}{\quad\text{or}\quad}{u}={\left\langle{1},{3}\right\rangle}{\quad\text{and}\quad}{v}={\left\langle-{2},{4}\right\rangle}$$
$$\displaystyle{u}\cdot{v}={\left\langle{1},{3}\right\rangle}\cdot{\left\langle-{2},{4}\right\rangle}=-{2}+{12}={10}$$
$$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{1}^{2}+{3}^{2}}}=\sqrt{{10}}$$
$$\displaystyle{\left|{\left|{v}\right|}\right|}=\sqrt{{{\left(-{2}\right)}^{2}+{4}^{2}}}=\sqrt{{20}}$$
Now plugging these into our equation, we find that
$$\displaystyle \cos{{\left(\theta\right)}}=\frac{10}{{\sqrt{{10}}\sqrt{{20}}}}=\frac{1}{\sqrt{{2}}}$$
Looking at the unit circle, we see that $$\displaystyle \cos{{\left(\theta\right)}}=\frac{1}{\sqrt{{2}}}\ {a}{t}\ {45}^{\circ}{\quad\text{and}\quad}{315}^{\circ}$$. Because the angle between two vectors must be between $$\displaystyle{0}^{\circ}{\quad\text{and}\quad}{180}^{\circ}$$, that means that the angle between vectors u and v is $$\displaystyle{45}^{\circ}$$