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Expert-verified Found in: Page 215 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find the angle${\mathbit{\theta }}$ between each of the pairs of vectors $\stackrel{\mathbf{\to }}{\mathbf{u}}$ and localid="1659433601917" $\stackrel{\mathbf{\to }}{\mathbf{v}}$ in exercises 4 through 6. $\stackrel{\mathbf{\to }}{\mathbf{u}}{\mathbf{=}}\left[\begin{array}{c}1\\ -1\\ 2\\ -2\end{array}],\stackrel{\to }{v}=\left[\begin{array}{c}2\\ 3\\ 4\\ 5\end{array}\right\right]$

The angle $\theta$ between $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ is about 97.4$°$.

See the step by step solution

## Step 1: Angle between two vectors

Consider two nonzero vectors $\stackrel{\mathbf{\to }}{\mathbf{x}}$ and $\stackrel{\mathbf{\to }}{\mathbf{y}}$ in ${{\mathbit{R}}}^{{\mathbf{n}}}$. The anglebetween these vectors

Is defined as:

${\mathbit{\theta }}{\mathbf{=}}{\mathbit{a}}{\mathbit{r}}{\mathbit{c}}{\mathbit{c}}{\mathbit{o}}{\mathbit{s}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}{\mathbf{=}}\frac{\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{\to }}{\mathbf{y}}}{\mathbf{|}\mathbf{|}\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{|}\mathbf{|}\mathbf{.}\mathbf{|}\mathbf{|}\stackrel{\mathbf{\to }}{\mathbf{y}}\mathbf{|}\mathbf{|}}$

## Step 2: Substitute the values into the angle formula

The given vectors are $\stackrel{\to }{u}=\left[\begin{array}{c}1\\ -1\\ 2\\ -2\end{array}\right]$ and $\stackrel{\to }{v}=\left[\begin{array}{c}2\\ 3\\ 4\\ 5\end{array}\right]$.

Now, calculate the angle by substituting the required values into the angle formula

$\theta =arccos\left(\mathrm{\theta }\right)=\frac{\stackrel{\to }{\mathrm{u}}.\stackrel{\to }{\mathrm{v}}}{||\stackrel{\to }{u}||.||\stackrel{\to }{v}||}$.

localid="1659428200763" $\theta =arccos\left(\mathrm{\theta }\right)=\frac{\stackrel{\to }{\mathrm{u}}.\stackrel{\to }{\mathrm{v}}}{||\stackrel{\to }{u}||.||\stackrel{\to }{v}||}\phantom{\rule{0ex}{0ex}}=arccos\left(\mathrm{\theta }\right)\frac{\left[\begin{array}{c}1\\ -1\\ 2\\ -2\end{array}\right].\left[\begin{array}{c}2\\ 3\\ 4\\ 5\end{array}\right]}{\left(\sqrt{{1}^{2}+{\left(-1\right)}^{2}+{2}^{2}+{\left(-2\right)}^{1}}\right)\left(\sqrt{{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}}\right)}\phantom{\rule{0ex}{0ex}}=arccos\left(\mathrm{\theta }\right)\frac{1.2-1.3+2.4-2.5}{\left(\sqrt{1+1+4+4}\right)\left(\sqrt{4+9+16+25}\right)}\phantom{\rule{0ex}{0ex}}=arccos\left(\mathrm{\theta }\right)\frac{2-3+8-10}{\sqrt{10.54}}\phantom{\rule{0ex}{0ex}}=arccos\left(\mathrm{\theta }\right)\frac{-3}{\sqrt{540}}\phantom{\rule{0ex}{0ex}}\approx 97.4°$

Hence, the value of $\theta$ is about $97.4°$. ### Want to see more solutions like these? 