Chapter 14.2, Problem 28WE

(a)

To determine

To find: the exponent form of the vector $\overrightarrow{u}$ and $\overrightarrow{v}$ .

Answer to Problem 28WE

The exponent form of the vectors are $\overrightarrow{u}=5.8\text{i}+1.55\text{j}$ and $\overrightarrow{v}=4.22\text{i}+9.06\text{j}$ .

Explanation of Solution

Given information:

The values are given by $\Vert u\Vert =6$ , $\Vert v\Vert =10$ , $\alpha =15°$ , and $\beta =65°$ .

The exponent form of the vector can be written as,

  $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ and $\overrightarrow{v}=\Vert v\Vert \cos \beta +\Vert v\Vert \sin \beta$ .

Substitute 6for $\Vert u\Vert$ , and $15°$ for $\alpha$ in the formula $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ to find the vector $\overrightarrow{u}$ .

  $\begin{array}{c}\overrightarrow{u}=6\cos 15+6\sin 15\\ =5.8\text{i}+1.55\text{j}\end{array}$

Substitute $10$ for $\Vert v\Vert$ , and $65°$ for $\beta$ in the formula $\overrightarrow{v}=\Vert v\Vert \cos \beta +\Vert v\Vert \sin \beta$ to find the vector $\overrightarrow{v}$ .

  $\begin{array}{c}\overrightarrow{v}=10\cos 65+10\sin 65\\ =4.22\text{i}+9.06\text{j}\end{array}$

Therefore, the exponent form of the vectors are $\overrightarrow{u}=5.8\text{i}+1.55\text{j}$ and $\overrightarrow{v}=4.22\text{i}+9.06\text{j}$ .

(b)

To determine

To find: the dot product of the vector $\overrightarrow{u}$ and $\overrightarrow{v}$ .

Answer to Problem 28WE

The dot product of the vectors is $38.57$ .

Explanation of Solution

Given information:

The vectors are given by are $\overrightarrow{u}=5.8\text{i}+1.55\text{j}$ and $\overrightarrow{v}=4.22\text{i}+9.06\text{j}$ .

Consider the vectors.

  $\overrightarrow{u}=5.8\text{i}+1.55\text{j}$

  $\overrightarrow{v}=4.22\text{i}+9.06\text{j}$

Find the dot product of the vectors.

  $\begin{array}{c}\overrightarrow{u}·\overrightarrow{v}=\left(5.8\text{i}+1.55\text{j}\right)·\left(4.22\text{i}+9.06\text{j}\right)\\ =\left(5.8\right)·\left(4.22\right)+\left(1.55\right)\left(9.06\right)\\ \approx 38.57\end{array}$

Therefore, the dot product of the vectors is $38.57$ .

(c)

To determine

To find: the exponent form of the vector $\overrightarrow{u}$ and $\overrightarrow{v}$ .

Answer to Problem 28WE

The dot product of the vectors is $38.57$ .

Explanation of Solution

Given information:

The values are given by $\Vert u\Vert =6$ , $\Vert v\Vert =10$ , $\alpha =15°$ , and $\beta =65°$ .

The dot product of the two vectors can be written as,

  $\overrightarrow{u}·\overrightarrow{v}=\Vert u\Vert \Vert v\Vert \cos \theta$ .

Find the value of $\theta$ .

  $\begin{array}{c}\theta =65-15\\ =50\end{array}$

Substitute 6for $\Vert u\Vert$ , $10$ for $\Vert v\Vert$ , and $50°$ for $\theta$ in the formula $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ to find the vector $\overrightarrow{u}$ .

  $\begin{array}{c}\overrightarrow{u}·\overrightarrow{v}=\left(6\right)\left(10\right)\cos 50°\\ =\left(60\right)0.0872\\ \approx 38.57\end{array}$

Therefore, the dot product of the vectors is $38.57$ .