Question

For each pair of vectors $\overrightarrow{\mathbf{u}}$and $\overrightarrow{v}$listed in Exercises 7 through 9, determine whether the angle $\mathit{\theta }$between $\overrightarrow{\mathbf{u}}$and$\overrightarrow{\mathbf{v}}$ is acute, obtuse, or right.

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

Short answer

The angle$\theta$ between$\overrightarrow{u}$ and$\overrightarrow{v}$ is acute.

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 angle between these vectors is defined as:

$\mathit{\theta }\mathbf{=}\mathit{a}\mathit{r}\mathit{c}\mathbf{cos}\mathbf{=}\frac{\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}}{\mathbf{\left|}\mathbf{\right|}\overrightarrow{\mathbf{x}}\mathbf{\left|}\mathbf{\right|}\mathbf{.}\mathbf{\left|}\mathbf{\right|}\overrightarrow{\mathbf{y}}\mathbf{\left|}\mathbf{\right|}}$

Find the angle

Now, find the angle by substituting the required values into the formula .

$\theta =arc\cos =\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\right|\overrightarrow{u}\left|\right|.\left|\right|\overrightarrow{\mathrm{v}}\left|\right|}$

$$\begin{gathered} \mathrm{\theta }=\arccos \left(\mathrm{\theta }\right)=\frac{\overrightarrow{\mathrm{u}}.\overrightarrow{\mathrm{v}}}{\left|\right|\overrightarrow{\mathrm{u}}\left|\right|.\left|\right|\overrightarrow{\mathrm{v}}\left|\right|} \\ =\arccos \left(\mathrm{\theta }\right)\frac{\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right].\left[\begin{array}{c}3\\ 4\\ 5\\ 3\end{array}\right]}{\left(\sqrt{1^2+{(-1)}^2+1^2+{(-1)}^2}\right)\left(\sqrt{3^2+4^2+5^2+3^2}\right)} \\ =\arccos \left(\mathrm{\theta }\right)\frac{1.3-1.4+1.5-1.3}{\left(\sqrt{1+1+1+1}\right)\left(\sqrt{9+16+25+9}\right)} \\ =\arccos \left(\mathrm{\theta }\right)\frac{3-4+5-3}{\sqrt{4.59}} \\ =\arccos \left(\mathrm{\theta }\right)\frac{1}{\sqrt{236}} \\ \approx 89.75 \end{gathered}$$

Hence, the value of angle$\theta$ is$89.75°$ and therefore acute.