Question

Find the angle $\mathit{\theta }$between each of the pairs of vectors $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}$and $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}$in exercises 4 through 6.

4. $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}\mathbf{=}\left[\begin{array}{c}1\\ 1\end{array}\right]\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\left[\begin{array}{c}7\\ 11\end{array}\right]\mathbf{}\mathbf{.}$

Short answer

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

Step-by-step solution

Angle between two vectors

Consider two nonzero vectors$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$androle="math" localid="1659434098948" $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{y}}$in ${\mathit{R}}^{\mathbf{n}}$.The angle between these vectors

Is defined as:

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

Substitute the values into the angle formula

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

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