Chapter 3, Problem 3P

(a)

To determine

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

Answer to Problem 3P

The $x$ component of $\overrightarrow{B}$ is $\underset{\_}{6.9}$ and $y$ component of $\overrightarrow{B}$ is $\underset{\_}{-1.7}$.

Explanation of Solution

Write the expression for the $x$ component of $\overrightarrow{B}$.

    ${\overrightarrow{B}}_x=B\cos \theta \widehat{x}$        (I)

Here, $B$ is the magnitude of vector $\overrightarrow{B}$, ${\overrightarrow{B}}_x$ is the $x$ component of $\overrightarrow{B}$, $\theta$ is the angle $\overrightarrow{B}$ makes with $x$ axis.

Write the expression for the $y$ component of $\overrightarrow{B}$

    ${\overrightarrow{B}}_y=B\sin \theta \widehat{y}$        (II)

Here, ${\overrightarrow{B}}_y$ is the $y$ component of $\overrightarrow{B}$.

Conclusion:

Substitute $7.1$ for $B$, $-14°$ for $\theta$ in equation (I) to find ${\overrightarrow{B}}_x$.

    $\begin{array}{c}{\overrightarrow{B}}_x=\left(7.1\right)\cos \left(-14°\right)\widehat{x}\\ =6.9\widehat{x}\end{array}$

Substitute $7.1$ for $B$, $-14°$ for $\theta$ in equation (II) to find ${\overrightarrow{B}}_y$.

    $\begin{array}{c}{\overrightarrow{B}}_y=\left(7.1\right)\sin \left(-14°\right)\widehat{y}\\ =-\left(7.1\right)\sin \left(14°\right)\widehat{y}\\ =-1.7\widehat{y}\end{array}$

Therefore, the $x$ component of $\overrightarrow{B}$ is $\underset{\_}{6.9}$ and $y$ component of $\overrightarrow{B}$ is $\underset{\_}{-1.7}$.

(b)

To determine

The magnitude and direction of $\overrightarrow{C}$.

Answer to Problem 3P

The magnitude $\overrightarrow{C}$ is $\underset{\_}{6.9}$ and direction of $\overrightarrow{C}$ is $\underset{\_}{105°}$ counter clockwise from the $x$ axis.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{C}$.

    $C=\sqrt{C_x^2+C_y^2}$        (III)

Here, $C$ is the magnitude of $\overrightarrow{C}$, $C_x$ is the magnitude of ${\overrightarrow{C}}_x$, $C_y$ is the magnitude of ${\overrightarrow{C}}_y$.

Write the expression for the $C_x$.

    $C_x=C\cos \theta$        (IV)

Use equation (IV) to solve for $\theta$.

    $\theta ={\cos }^{-1}\left(\frac{C_x}{C}\right)$        (V)

Conclusion:

Substitute $-1.8$ for $C_x$, $-6.7$ for $C_y$ in equation (III) to find $C$.

    $\begin{array}{c}C=\sqrt{{\left(-1.8\right)}^2+{\left(-6.7\right)}^2}\\ =6.9\end{array}$

Substitute $-1.8$ for $C_x$, $6.9$ for $C$ in equation (V) to find $\theta$.

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-1.8}{6.9}\right)\\ =105°\end{array}$

Counter clockwise direction from the $x$ axis.

Therefore, the magnitude $\overrightarrow{C}$ is $\underset{\_}{6.9}$ and direction of $\overrightarrow{C}$ is $\underset{\_}{105°}$ counter clockwise from the $x$ axis.

(c)

To determine

The magnitude $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{10}$ and direction of $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{150°}$.

Answer to Problem 3P

The magnitude and direction of $\overrightarrow{C}-\overrightarrow{B}$ is .

Explanation of Solution

Write the expression for the $\overrightarrow{C}-\overrightarrow{B}$.

    $\begin{array}{c}\overrightarrow{C}-\overrightarrow{B}=\left(C_x\widehat{x}+C_y\widehat{y}\right)-\left(B_x\widehat{x}+B_y\widehat{y}\right)\\ =\left(C_x-B_x\right)\widehat{x}+\left(C_y-B_y\right)\widehat{y}\end{array}$        (VI)

Write the expression for the direction of the $\overrightarrow{C}-\overrightarrow{B}$.

    $\overrightarrow{C}-\overrightarrow{B}=\sqrt{{\left(C_x-B_x\right)}^2+{\left(C_y-B_y\right)}^2}$        (VII)

Write the expression for the direction of $\overrightarrow{C}-\overrightarrow{B}$.

    $\theta ={\cos }^{-1}\left[\frac{{\left(\overrightarrow{C}-\overrightarrow{B}\right)}_x}{\left(\overrightarrow{C}-\overrightarrow{B}\right)}\right]$        (VIII)

Conclusion:

Substitute $-1.8$ for $C_x$, $-6.7$ for $C_y$, $6.9$ for $B_x$, $-1.7$ for $B_y$ in equation (VI) to find $\overrightarrow{C}-\overrightarrow{B}$.

    $\begin{array}{c}\left|\overrightarrow{C}-\overrightarrow{B}\right|=\sqrt{{\left(-1.8-6.9\right)}^2+{\left(-6.7-\left(-1.7\right)\right)}^2}\\ =\sqrt{{\left(-8.7\right)}^2+{\left(-5\right)}^2}\\ =10\end{array}$

Substitute $-8.7$ for ${\left(\overrightarrow{C}-\overrightarrow{B}\right)}_x$, $10$ for $\left(\overrightarrow{C}-\overrightarrow{B}\right)$ in equation (VIII) to find $\theta$.

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-8.7}{10}\right)\\ =150°\end{array}$

Counter clockwise from the $x$ axis.

Therefore, the magnitude $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{10}$ and direction of $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{150°}$.

(d)

To determine

The $x$ component of $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{-8.6}$ and $y$ component of $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{5}$.

Answer to Problem 3P

The $x$ and $y$ components of $\overrightarrow{C}-\overrightarrow{B}$ is .

Explanation of Solution

Write the expression for the $x$ component of $\overrightarrow{C}-\overrightarrow{B}$.

    ${\left(\overrightarrow{C}-\overrightarrow{B}\right)}_x=\left(\overrightarrow{C}-\overrightarrow{B}\right)\cos \theta \widehat{x}$        (IX)

Write the expression for the $y$ component of $\overrightarrow{C}-\overrightarrow{B}$.

    ${\left(\overrightarrow{C}-\overrightarrow{B}\right)}_y=\left(\overrightarrow{C}-\overrightarrow{B}\right)\sin \theta \widehat{y}$        (X)

Conclusion:

Substitute $10$ for $\overrightarrow{C}-\overrightarrow{B}$, $150°$ for $\varphi$ in equation (IX) to find $x$ component of $\overrightarrow{C}-\overrightarrow{B}$.

    $\begin{array}{c}{\left(\overrightarrow{C}-\overrightarrow{B}\right)}_x=\left(10\right)\cos \left(150°\right)\widehat{x}\\ =-8.6\widehat{x}\end{array}$

Substitute $10$ for $\overrightarrow{C}-\overrightarrow{B}$, $150°$ for $\varphi$ in equation (X) to find $y$ component of $\overrightarrow{C}-\overrightarrow{B}$.

    $\begin{array}{c}{\left(\overrightarrow{C}-\overrightarrow{B}\right)}_y=\left(10\right)\sin \left(150°\right)\widehat{y}\\ =5\widehat{y}\end{array}$

Therefore, the $x$ component of $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{-8.6}$ and $y$ component of $\overrightarrow{C}-\overrightarrow{B}$ is $\underset{\_}{5}$.