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### 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 $\stackrel{\mathbf{⇀}}{\mathbf{u}}$ and $\stackrel{\mathbf{⇀}}{\mathbf{v}}$ in exercises 4 through 6.4. $\stackrel{\mathbf{⇀}}{\mathbf{u}}{\mathbf{=}}\left[\begin{array}{c}1\\ 1\end{array}\right]{\mathbf{,}}\stackrel{\mathbf{⇀}}{\mathbf{v}}{\mathbf{=}}\left[\begin{array}{c}7\\ 11\end{array}\right]{\mathbf{}}{\mathbf{.}}$

The angle $\theta$ between $\stackrel{⇀}{u}$ and $\stackrel{⇀}{v}$ is about $12.58°$ .

See the step by step solution

## Step 1: Angle between two vectors

Consider two nonzero vectors $\stackrel{\mathbf{⇀}}{\mathbf{x}}$ and role="math" localid="1659434098948" $\stackrel{\mathbf{⇀}}{\mathbf{y}}$ in ${{\mathbit{R}}}^{{\mathbf{n}}}$. The angle between these vectors

Is defined as:

role="math" localid="1659434271678" ${\mathbit{\theta }}{\mathbf{=}}{\mathbit{a}}{\mathbit{r}}{\mathbit{c}}{\mathbit{c}}{\mathbit{o}}{\mathbit{s}}\frac{\stackrel{\mathbf{⇀}}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{⇀}}{\mathbf{y}}}{\mathbf{||}\stackrel{\mathbf{⇀}}{\mathbf{x}}\mathbf{||}\mathbf{.}\mathbf{||}\stackrel{\mathbf{⇀}}{\mathbf{y}}\mathbf{||}}$

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

$\theta =arccos\frac{\stackrel{⇀}{u}.\stackrel{⇀}{v}}{||\stackrel{⇀}{u}||.||\stackrel{⇀}{v}||}\phantom{\rule{0ex}{0ex}}=arccos\frac{\left[\left[11\right]\left[\begin{array}{c}7\\ 11\end{array}\right]\right]}{\left(\sqrt{1-1+1-1}\right)\left(\sqrt{7-7+11-11}\right)}\phantom{\rule{0ex}{0ex}}=arccos\frac{1-7+1-11}{\left(\sqrt{1+1}\right)\left(\sqrt{49+121}\right)}\phantom{\rule{0ex}{0ex}}=arccos\frac{18}{\sqrt{340}}\phantom{\rule{0ex}{0ex}}=arccos\frac{18}{18.44}\phantom{\rule{0ex}{0ex}}=arccos\left(0.976\right)\phantom{\rule{0ex}{0ex}}\approx 12.58°$

Hence, the value of $\theta$ is about $12.58°$ .