Chapter 1, Problem 38P

(a)

To determine

Draw the vector sum $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ and the vector difference $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$.

Answer to Problem 38P

The vector sum $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ and the vector difference $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$ is drawn in figure 1.

Explanation of Solution

The figure 1 shows the vector sum of $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ and the vector difference of $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$.

Conclusion:

Therefore, the vector sum $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ and the vector difference $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$ is drawn.

(b)

To determine

The components of the unit vectors for $\overrightarrow{C}$ and $\overrightarrow{D}$.

Answer to Problem 38P

The components of the unit vectors for $\overrightarrow{C}$ is $5.00\widehat{i}+4.00\widehat{j}$ and $\overrightarrow{D}$ is $-1.00\widehat{i}+8.00\widehat{j}$.

Explanation of Solution

Write the expression for the given vector sum $\overrightarrow{C}$.

  $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$        (I)

Write the expression for the given vector difference $\overrightarrow{D}$.

  $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$        (II)

Conclusion:

Substitute $\left(2.00\widehat{i}+6.00\widehat{j}\right)$ for $\overrightarrow{A}$ and $\left(3.00\widehat{i}-2.00\widehat{j}\right)$ for $\overrightarrow{B}$ in the equation (I) to find $\overrightarrow{C}$.

  $\begin{array}{c}\overrightarrow{C}=\left(2.00\widehat{i}+6.00\widehat{j}\right)+\left(3.00\widehat{i}-2.00\widehat{j}\right)\\ =5.00\widehat{i}+4.00\widehat{j}\end{array}$

Substitute $\left(2.00\widehat{i}+6.00\widehat{j}\right)$ for $\overrightarrow{A}$ and $\left(3.00\widehat{i}-2.00\widehat{j}\right)$ for $\overrightarrow{B}$ in the equation (II) to find $\overrightarrow{D}$.

  $\begin{array}{c}\overrightarrow{D}=\left(2.00\widehat{i}+6.00\widehat{j}\right)-\left(3.00\widehat{i}-2.00\widehat{j}\right)\\ =-1.00\widehat{i}+8.00\widehat{j}\end{array}$

Therefore, the components of the unit vectors for $\overrightarrow{C}$ is $5.00\widehat{i}+4.00\widehat{j}$ and $\overrightarrow{D}$ is $-1.00\widehat{i}+8.00\widehat{j}$.

(c)

To determine

The magnitude and direction of the $\overrightarrow{C}$ and $\overrightarrow{D}$ with respect to the positive x axis.

Answer to Problem 38P

The magnitude and direction of the $\overrightarrow{C}$ is $6.40$ at $38.7°$, magnitude and direction of the $\overrightarrow{D}$ is $8.06$ at $97.1°$ with respect to the positive x axis.

Explanation of Solution

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

  $\left|\overrightarrow{C}\right|=\sqrt{{\text{C}}_x^2+{\text{C}}_y^2}$        (III)

Here, $\left|\overrightarrow{C}\right|$ is the magnitude of the vector, ${\text{C}}_x$ and ${\text{C}}_y$ are the x and y components of the vector.

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

  $\theta ={\tan }^{-1}\left(\frac{{\text{C}}_y}{{\text{C}}_x}\right)$        (IV)

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

  $\left|\overrightarrow{D}\right|=\sqrt{{\text{D}}_x^2+{\text{D}}_y^2}$        (V)

Here, $\left|\overrightarrow{D}\right|$ is the magnitude of the vector, ${\text{D}}_x$ and ${\text{D}}_y$ are the x and y components of the vector.

Write the expression for the direction of the vector $\overrightarrow{D}$.

  $\theta ={\tan }^{-1}\left(\frac{{\text{D}}_y}{{\text{D}}_x}\right)$        (VI)

Conclusion:

Substitute $5.00\widehat{i}$ for ${\text{C}}_x$ and $4.00\widehat{j}$ for ${\text{C}}_y$ in equation (III) to find $\left|\overrightarrow{C}\right|$.

  $\begin{array}{c}\left|\overrightarrow{C}\right|=\sqrt{{\left(5.00\right)}^2+{\left(4.00\right)}^2}\\ =6.40\end{array}$

Substitute $5.00\widehat{i}$ for ${\text{C}}_x$ and $4.00\widehat{j}$ for ${\text{C}}_y$ in equation (IV) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{4.00}{5.00}\right)\\ =38.7°\end{array}$

Substitute $-1.00\widehat{i}$ for ${\text{D}}_x$ and $8.00\widehat{j}$ for ${\text{D}}_y$ in equation (V) to find $\left|\overrightarrow{D}\right|$.

  $\begin{array}{c}\left|\overrightarrow{D}\right|=\sqrt{{\left(-1.00\right)}^2+{\left(8.00\right)}^2}\\ =8.06\end{array}$

Substitute $-1.00\widehat{i}$ for ${\text{D}}_x$ and $8.00\widehat{j}$ for ${\text{D}}_y$ in equation (VI) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{8.00}{-1.00}\right)\\ =82.9°\\ \theta =180°-82.9°\\ =97.1°\end{array}$

Therefore, the magnitude and direction of the $\overrightarrow{C}$ is $6.40$ at $38.7°$, magnitude and direction of the $\overrightarrow{D}$ is $8.06$ at $97.1°$ with respect to the positive x axis.