Chapter 14.2, Problem 29WE

(a)

To determine

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

Answer to Problem 29WE

The exponent form of the vectors are $\overrightarrow{u}=4.68\text{i}+11.58\text{j}$ and $\overrightarrow{v}=-7.89\text{i}+16.17\text{j}$ .

Explanation of Solution

Given information:

The values are given by $\Vert u\Vert =12.5$ , $\Vert v\Vert =18.0$ , $\alpha =68°$ , and $\beta =116°$ .

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 12.5for $\Vert u\Vert$ , and $68°$ for $\alpha$ in the formula $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ to find the vector $\overrightarrow{u}$ .

  $\begin{array}{c}\overrightarrow{u}=12.5\cos 68+12.5\sin 68\\ =4.68\text{i}+11.58\text{j}\end{array}$

Substitute $18$ for $\Vert v\Vert$ , and $116°$ for $\beta$ in the formula $\overrightarrow{v}=\Vert v\Vert \cos \beta +\Vert v\Vert \sin \beta$ to find the vector $\overrightarrow{v}$ .

  $\begin{array}{c}\overrightarrow{v}=18\cos 116+18\sin 116\\ =-7.89\text{i}+16.17\text{j}\end{array}$

Therefore, the exponent form of the vectors are $\overrightarrow{u}=4.68\text{i}+11.58\text{j}$ and $\overrightarrow{v}=-7.89\text{i}+16.17\text{j}$ .

(b)

To determine

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

Answer to Problem 29WE

The dot product of the vectors is $150.32$ .

Explanation of Solution

Given information:

The vectors are given by are $\overrightarrow{u}=4.68\text{i}+11.58\text{j}$ and $\overrightarrow{v}=-7.89\text{i}+16.17\text{j}$

Consider the vectors.

  $\overrightarrow{u}=4.68\text{i}+11.58\text{j}$

  $\overrightarrow{v}=-7.89\text{i}+16.17\text{j}$

Find the dot product of the vectors.

  $\begin{array}{c}\overrightarrow{u}·\overrightarrow{v}=\left(4.681\text{i}+11.58\text{j}\right)·\left(-7.89\text{i}+16.17\text{j}\right)\\ =\left(4.68\right)·\left(-7.89\right)+\left(11.58\right)\left(16.17\right)\\ \approx 150.32\end{array}$

Therefore, the dot product of the vectors is $150.32$ .

(c)

To determine

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

Answer to Problem 29WE

The dot product of the vectors is $150.32$ .

Explanation of Solution

Given information:

The values are given by $\Vert u\Vert =12.5$ , $\Vert v\Vert =18.0$ , $\alpha =68°$ , and $\beta =116°$ .

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 $\theta$ .

  $\begin{array}{c}\theta =116-68\\ =48\end{array}$

Substitute 12.5for $\Vert u\Vert$ , $18.0$ for $\Vert v\Vert$ , and $48°$ for $\theta$ in the formula $\overrightarrow{u}=\Vert u\Vert \cos \alpha +\Vert u\Vert \sin \alpha$ to find the vector $\overrightarrow{u}$ .

  $\begin{array}{c}\overrightarrow{u}·\overrightarrow{v}=\left(12.5\right)\left(18.0\right)\cos 48°\\ =\left(225\right)\cos 48°\\ \approx 150.55\end{array}$

Therefore, the dot product of the vectors is $150.32$ .