Question

Find the angle$\mathit{\theta }$between each of the pairs of vectors $\overrightarrow{\mathbf{u}}$and localid="1659433601917" $\overrightarrow{\mathbf{v}}$in exercises 4 through

6.$\overrightarrow{\mathbf{u}}\mathbf{=}\left[\left.\begin{array}{c}1\\ -1\\ 2\\ -2\end{array}\right],\overrightarrow{v}=\left[\begin{array}{c}2\\ 3\\ 4\\ 5\end{array}\right.\right]$

Short answer

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

Step-by-step solution

Angle between two vectors

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

Is defined as:

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

Substitute the values into the angle formula

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

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

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

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

Hence, the value of$\theta$is about $97.4°$.