Chapter 3, Problem 23P

(a)

To determine

The value of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

Answer to Problem 23P

$\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ is $\text{2}\widehat{\text{i}}-\text{6}\widehat{\text{j}}$.

Explanation of Solution

Write the equation to find sum of two vectors in component form.

    $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\text{=}\left(A_x\widehat{i}+A_y\widehat{j}\right)+\left(B_x\widehat{i}+B_y\widehat{j}\right)$

Here, $A_x$ is the x-component of $\overrightarrow{\text{A}}$, $B_x$ is the x-component of $\overrightarrow{\text{B}}$, $A_y$ is the y-component of $\overrightarrow{\text{A}}$, and $B_y$ is the y-component of $\overrightarrow{\text{B}}$.

Rearrange the above expression.

  $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\text{=}\left(A_x+B_x\right)\widehat{i}+\left(A_y+B_y\right)\widehat{j}$

Conclusion:

Substitute $3$ for $A_x$, $-1$ for $B_x$, $-2$ for $A_y$, and $-4$ for $B_y$ in the equation to find $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

    $\begin{array}{c}\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\text{=}\left(3+\left(-1\right)\right)\widehat{i}+\left(-2+\left(-4\right)\right)\widehat{j}\\ =\text{2}\widehat{\text{i}}-\text{6}\widehat{\text{j}}\end{array}$

Therefore, $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ is $\text{2}\widehat{\text{i}}-\text{6}\widehat{\text{j}}$.

(b)

To determine

The value of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

Answer to Problem 23P

$\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ is $\text{4}\widehat{\text{i}}\text{+2}\widehat{\text{j}}$.

Explanation of Solution

Write the equation to find the difference of two vectors in component form.

    $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\text{=}\left(A_x\widehat{i}+A_y\widehat{j}\right)-\left(B_x\widehat{i}+B_y\widehat{j}\right)$

Rearrange the above expression.

  $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\text{=}\left(A_x-B_x\right)\widehat{i}+\left(A_y-B_y\right)\widehat{j}$

Conclusion:

Substitute $3$ for $A_x$, $-1$ for $B_x$, $-2$ for $A_y$, and $-4$ for $B_y$ in the equation to find $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

    $\begin{array}{c}\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\text{=}\left(3-\left(-1\right)\right)\widehat{i}+\left(-2-\left(-4\right)\right)\widehat{j}\\ =4\widehat{\text{i}}\text{+}2\widehat{\text{j}}\end{array}$

Therefore, $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ is $\text{4}\widehat{\text{i}}\text{+2}\widehat{\text{j}}$.

(c)

To determine

The magnitude of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

Answer to Problem 23P

The magnitude of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ is $6.32$.

Explanation of Solution

Write the equation to find the magnitude of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

    $\left|\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\right|=\sqrt{x_{ab}^2+y_{ab}^2}$

Here, $\left|\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\right|$ is the magnitude of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$, $x_{ab}$ is the x-component of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$, and $y_{ab}$ is the y-component of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

Conclusion:

Substitute $2$ for $x_{ab}$ and $-6$ for $y_{ab}$ in the equation to find $\left|\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\right|$.

    $\begin{array}{c}\left|\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\right|=\sqrt{2^2+{\left(-6\right)}^2}\\ =\sqrt{40}\\ =6.32\end{array}$

Therefore, the magnitude of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ is $6.32$.

(d)

To determine

The magnitude of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

Answer to Problem 23P

The magnitude of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$ is $4.47$.

Explanation of Solution

Write the equation to find the magnitude of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

    $\left|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\right|=\sqrt{x_{ab}^2+y_{ab}^2}$

Here, $\left|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\right|$ is the magnitude of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$, $x_{ab}$ is the x-component of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$, and $y_{ab}$ is the y-component of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

Conclusion:

Substitute $4$ for $x_{ab}$ and $2$ for $y_{ab}$ in the equation to find $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

    $\begin{array}{c}\left|\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\right|=\sqrt{4^2+2^2}\\ =\sqrt{20}\\ =4.47\end{array}$

Therefore, the magnitude of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$ is $4.47$.

(d)

To determine

The directions of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ and $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

Answer to Problem 23P

The directions of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ and $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$ are $-71.6°$ and $26.6°$ respectively.

Explanation of Solution

Write the equation to find the direction of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ or $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

  $\theta ={\tan }^{-1}\left(\frac{y_{ab}}{x_{ab}}\right)$

Here, $\theta$ is the angle made by $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ or $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$ with positive x-axis.

Conclusion:

Substitute $-6$ for $y_{ab}$ and $2$ for $x_{ab}$ in the above equation to find $\theta$ of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-6}{2}\right)\\ =-71.6°\end{array}$

The negative sign indicates the angle made by $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ with positive x-axis is measured in clockwise direction.

Substitute $4$ for $y_{ab}$ and $2$ for $x_{ab}$ in the above equation to find $\theta$ of $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{4}{2}\right)\\ =26.6°\end{array}$

Therefore, the directions of $\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}$ and $\overrightarrow{\text{A}}-\overrightarrow{\text{B}}$ are $-71.6°$ and $26.6°$ respectively.