Chapter 3, Problem 20P

(a)

To determine

The magnitude and direction $\overrightarrow{A}+\overrightarrow{B}$.

Answer to Problem 20P

The magnitude of $\overrightarrow{A}+\overrightarrow{B}$ is $\underset{\_}{\left|\overrightarrow{A}+\overrightarrow{B}\right|=2.0\text{units}}$, at $60°$ CW from $-x$ axis.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{A}+\overrightarrow{B}$

$\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{{[{\left(\overrightarrow{A}+\overrightarrow{B}\right)}_x]}^2+{[{\left(\overrightarrow{A}+\overrightarrow{B}\right)}_y]}^2}$ (I)

Write the expression for direction of $\overrightarrow{A}+\overrightarrow{B}$

$\theta ={\tan }^{-1}\frac{\text{opposite}}{\text{adjacent}}$ (II)

Conclusion:

Substitute $-1.0$ for ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_x$, $\sqrt{3.0}$ for ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_y$ in equation (I)

$\begin{array}{c}\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{{[-1.0]}^2+{[3.0]}^2}\\ =2.0\text{units}\end{array}$

Substitute $\sqrt{3.0}$ for opposite side, and $1.0$ for adjacent side in equation (II)

$\begin{array}{c}\theta ={\tan }^{-1}\frac{\sqrt{3.0}}{1.0}\\ =60°\end{array}$

Therefore the magnitude of $\overrightarrow{A}+\overrightarrow{B}$ is $\underset{\_}{\left|\overrightarrow{A}+\overrightarrow{B}\right|=2.0\text{units}}$, at $60°$ CW from $-x$ axis.

(b)

To determine

The magnitude and direction $\overrightarrow{A}-\overrightarrow{B}$.

Answer to Problem 20P

The magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is $\underset{\_}{\left|\overrightarrow{A}-\overrightarrow{B}\right|=2.0\text{units}}$, at $60°$ CW from $+x$ axis.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{A}-\overrightarrow{B}$

$\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt{{[{\left(\overrightarrow{A}-\overrightarrow{B}\right)}_x]}^2+{[{\left(\overrightarrow{A}-\overrightarrow{B}\right)}_y]}^2}$ (I)

Write the expression for direction of $\overrightarrow{A}+\overrightarrow{B}$

$\theta ={\tan }^{-1}\frac{\text{opposite}}{\text{adjacent}}$ (II)

Conclusion:

Substitute $1.0$ for ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_x$, $\sqrt{3.0}$ for ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_y$ in equation (I)

$\begin{array}{c}\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{{[1.0]}^2+{[\sqrt{3.0}]}^2}\\ =2.0\text{units}\end{array}$

Substitute $\sqrt{3.0}$ for opposite side, and $1.0$ for adjacent side in equation (II)

$\begin{array}{c}\theta ={\tan }^{-1}\frac{\sqrt{3.0}}{1.0}\\ =60°\end{array}$

Therefore, the magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is $\underset{\_}{\left|\overrightarrow{A}-\overrightarrow{B}\right|=2.0\text{units}}$, at $60°$ CW from $+x$ axis.

(c)

To determine

The $x$ and $y$ component of $\overrightarrow{\text{B}}-\overrightarrow{\text{A}}$.

Answer to Problem 20P

The $x$ component is $\underset{\_}{B_x=-1.0\text{units}}$, and $y$ component is $\underset{\_}{A_y=-\sqrt{3.0}\text{units}}$.

Explanation of Solution

Consider the figure given below representing individual vectors as well as resultant vectors.

The resultant vector $\overrightarrow{\text{B}}-\overrightarrow{\text{A}}$ lies in the third quadrant can be resolved into two components $B_x$ and $A_y$(since, $\overrightarrow{A}$ in the positive $y$ axis has no $x$ component, and $\overrightarrow{B}$ in negative $x$ axis has no $y$ component). The magnitude of $B_x=-1.0\text{units}$, and $A_y$ is, $A_y=-\sqrt{3.0}\text{units}$.

Conclusion:

Therefore, the $x$ component is $\underset{\_}{B_x=-1.0\text{units}}$, and $y$ component is $\underset{\_}{A_y=-\sqrt{3.0}\text{units}}$