Chapter 14.2, Problem 30WE

(a)

To determine

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

Answer to Problem 30WE

The exponent form of the vectors are $\overrightarrow{u}=4.74\text{i}+7.3\text{j}$ and $\overrightarrow{v}=5.83\text{i}-3.10\text{j}$ .

Explanation of Solution

Given information:

The values are given by $\Vert u\Vert =8.70$ , $\Vert v\Vert =6.60$ , $\alpha =57°$ , and $\beta =-28°$ .

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 8.70 for $\Vert u\Vert$ , and $57°$ for $\alpha$ in the formula $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ to find the vector $\overrightarrow{u}$ .

  $\begin{array}{c}\overrightarrow{u}=8.70\cos 57+8.70\sin 57\\ =4.74\text{i}+7.3\text{j}\end{array}$

Substitute $6.60$ for $\Vert v\Vert$ , and $-28°$ for $\beta$ in the formula $\overrightarrow{v}=\Vert v\Vert \cos \beta +\Vert v\Vert \sin \beta$ to find the vector $\overrightarrow{v}$ .

  $\begin{array}{c}\overrightarrow{v}=6.60\cos \left(-28°\right)+6.60\sin \left(-28°\right)\\ =5.83\text{i}-3.10\text{j}\end{array}$

Therefore, the exponent form of the vectors are $\overrightarrow{u}=4.74\text{i}+7.3\text{j}$ and $\overrightarrow{v}=5.83\text{i}-3.10\text{j}$ .

(b)

To determine

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

Answer to Problem 30WE

The dot product of the vectors is $5$ .

Explanation of Solution

Given information:

The vectors are given by $\overrightarrow{u}=4.74\text{i}+7.3\text{j}$ and $\overrightarrow{v}=5.83\text{i}-3.10\text{j}$ .

Consider the vectors.

  $\overrightarrow{u}=4.74\text{i}+7.3\text{j}$

  $\overrightarrow{v}=5.83\text{i}-3.10\text{j}$

Find the dot product of the vectors.

  $\begin{array}{c}\overrightarrow{u}·\overrightarrow{v}=\left(4.74\text{i}+7.3\text{j}\right)·\left(5.83\text{i}-3.10\text{j}\right)\\ =\left(4.74\right)·\left(5.83\right)+\left(7.3\right)\left(-3.10\right)\\ =27.63-22.63\\ \approx 5\end{array}$

Therefore, the dot product of the vectors is $5$ .

(c)

To determine

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

Answer to Problem 30WE

The dot product of the vectors is $5$ .

Explanation of Solution

Given information:

The values are given by $\Vert u\Vert =8.70$ , $\Vert v\Vert =6.60$ , $\alpha =57°$ , and $\beta =-28°$ .

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 $\theta$ .

  $\begin{array}{c}\theta =57-\left(-28\right)\\ =57+28\\ =85°\end{array}$

Substitute 8.70 for $\Vert u\Vert$ , $6.60$ for $\Vert v\Vert$ , and $85°$ for $\theta$ in the formula $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ to find the vector $\overrightarrow{u}$ .

  $\begin{array}{c}\overrightarrow{u}·\overrightarrow{v}=\left(8.70\right)\left(6.60\right)\cos 85°\\ =\left(57.42\right)0.0872\\ \approx 5\end{array}$

Therefore, the dot product of the vectors is $5$ .