Question

Find the angle between the vectors displaystyle<{1},{3}>{quadtext{and}quad}<-{2},{4}>

Vectors and spaces
ANSWERED
asked 2021-01-02
Find the angle between the vectors \(\displaystyle<{1},{3}>{\quad\text{and}\quad}<-{2},{4}>\)

Answers (1)

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 \(\displaystyle{45}^{\circ}\)
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