Chapter 3, Problem 60PQ

(a)

To determine

The vector $\overrightarrow{A}+\overrightarrow{B}$

Answer to Problem 60PQ

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

Explanation of Solution

Write an expression for the vector $\overrightarrow{A}$.

    $\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}$ (I)

Here, $\overrightarrow{A}$ is the vector, $A_x$ is the horizontal component of the vector and $A_y$ is the vertical component of the vector $\overrightarrow{A}$.

Write an expression for the vector $\overrightarrow{B}$.

    $\overrightarrow{B}=B_x\widehat{i}+B_y\widehat{j}$ (II)

Here, $\overrightarrow{B}$ is the vector, $B_x$ is the horizontal component of the vector and $B_y$ is the vertical component of the vector $\overrightarrow{B}$.

Write an expression for $\overrightarrow{A}+\overrightarrow{B}$.

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

Here, $\overrightarrow{A}+\overrightarrow{B}$ is the sum of the vectors.

Conclusion:

The vectors are $\overrightarrow{A}=-4\overrightarrow{i}+3\widehat{j}$ and $\overrightarrow{B}=-2\widehat{i}-5\widehat{j}$.

Substitute $-4$ for $A_x$, $3$ for $A_y$, $-2$ for $B_x$ and $-5$ for $B_y$ in equation (III) to find $\overrightarrow{A}+\overrightarrow{B}$.

    $\begin{array}{c}\overrightarrow{A}+\overrightarrow{B}=\left(-4+-2\right)\widehat{i}+\left(3+-5\right)\widehat{j}\\ =\left(-6\right)\widehat{i}+\left(-2\right)\widehat{j}\end{array}$

Thus, the vector $\overrightarrow{A}+\overrightarrow{B}$ is $-6\widehat{i}-2\widehat{j}$.

(b)

To determine

The vector $\overrightarrow{A}-\overrightarrow{B}$

Answer to Problem 60PQ

The vector $\overrightarrow{A}-\overrightarrow{B}$ is $-2\widehat{i}+8\widehat{j}$.

Explanation of Solution

Write an expression for $\overrightarrow{A}-\overrightarrow{B}$.

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

Here, $\overrightarrow{A}-\overrightarrow{B}$ is the difference of the vectors.

Conclusion:

Substitute $-4$ for $A_x$, $3$ for $A_y$, $-2$ for $B_x$ and $-5$ for $B_y$ in equation (IV) to find $\overrightarrow{A}-\overrightarrow{B}$.

    $\begin{array}{c}\overrightarrow{A}+\overrightarrow{B}=\left(-4-\left(-2\right)\right)\widehat{i}+\left(3+-\left(-5\right)\right)\widehat{j}\\ =\left(-2\right)\widehat{i}+\left(8\right)\widehat{j}\end{array}$

Thus, the vector $\overrightarrow{A}+\overrightarrow{B}$ is $-2\widehat{i}+8\widehat{j}$.

(c)

To determine

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

Answer to Problem 60PQ

The magnitude of $\overrightarrow{A}+\overrightarrow{B}$ is $6.3$.

Explanation of Solution

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

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

Here, $\left|\overrightarrow{A}+\overrightarrow{B}\right|$ is the magnitude of $\overrightarrow{A}+\overrightarrow{B}$, ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_x$ is the x component of $\overrightarrow{A}+\overrightarrow{B}$ and ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_y$ is the y component of $\overrightarrow{A}+\overrightarrow{B}$

Conclusion:

Substitute $-6$ for ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_x$ and $-2$ for ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_y$ in equation (V) to find $\left|\overrightarrow{A}+\overrightarrow{B}\right|$.

    $\begin{array}{c}\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{{\left(-6\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{36+4}\\ =6.3\end{array}$

Thus, the magnitude of $\overrightarrow{A}+\overrightarrow{B}$ is $6.3$.

(d)

To determine

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

Answer to Problem 60PQ

The magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is $8.2$.

Explanation of Solution

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

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

Here, $\left|\overrightarrow{A}-\overrightarrow{B}\right|$ is the magnitude of $\overrightarrow{A}-\overrightarrow{B}$, ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_x$ is the x component of $\overrightarrow{A}-\overrightarrow{B}$ and ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_y$ is the y component of $\overrightarrow{A}-\overrightarrow{B}$

Conclusion:

Substitute $-2$ for ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_x$ and $8$ for ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_y$ in equation (VI) to find $\left|\overrightarrow{A}-\overrightarrow{B}\right|$.

    $\begin{array}{c}\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt{{\left(-2\right)}^2+{\left(8\right)}^2}\\ =\sqrt{4+64}\\ =8.2\end{array}$

Thus, the magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is $8.2$.

(e)

To determine

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

Answer to Problem 60PQ

The direction of $\overrightarrow{A}+\overrightarrow{B}$ is $18.4°\text{below the -x axis}$ and direction of $\overrightarrow{A}-\overrightarrow{B}$ is $76.0°\text{above the -x axis}$.

Explanation of Solution

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

    ${\theta }_{\overrightarrow{A}+\overrightarrow{B}}={\tan }^{-1}\frac{{\left(\overrightarrow{A}+\overrightarrow{B}\right)}_y}{{\left(\overrightarrow{A}+\overrightarrow{B}\right)}_x}$ (VII)

Here, ${\theta }_{\overrightarrow{A}+\overrightarrow{B}}$ is the direction of $\overrightarrow{A}+\overrightarrow{B}$.

Write an expression for direction of $\overrightarrow{A}-\overrightarrow{B}$.

    ${\theta }_{\overrightarrow{A}-\overrightarrow{B}}={\tan }^{-1}\frac{{\left(\overrightarrow{A}-\overrightarrow{B}\right)}_y}{{\left(\overrightarrow{A}-\overrightarrow{B}\right)}_x}$ (VIII)

Here, ${\theta }_{\overrightarrow{A}-\overrightarrow{B}}$ is the direction of $\overrightarrow{A}-\overrightarrow{B}$.

Conclusion:

Substitute $-6$ for ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_x$ and $-2$ for ${\left(\overrightarrow{A}+\overrightarrow{B}\right)}_y$ in equation (VII) to find ${\theta }_{\overrightarrow{A}+\overrightarrow{B}}$.

    $\begin{array}{c}{\theta }_{\overrightarrow{A}+\overrightarrow{B}}={\tan }^{-1}\left(\frac{-2}{-6}\right)\\ ={\tan }^{-1}\left(\frac{1}{3}\right)\\ =18.4°\text{below the -x axis}\end{array}$

Substitute $-2$ for ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_x$ and $8$ for ${\left(\overrightarrow{A}-\overrightarrow{B}\right)}_y$ in equation (VIII) to find ${\theta }_{\overrightarrow{A}-\overrightarrow{B}}$.

    $\begin{array}{c}{\theta }_{\overrightarrow{A}-\overrightarrow{B}}={\tan }^{-1}\left(\frac{8}{-2}\right)\\ ={\tan }^{-1}\left(-4\right)\\ =76.0°\text{above the -x axis}\end{array}$

Thus, the direction of $\overrightarrow{A}+\overrightarrow{B}$ is $18.4°\text{below the -x axis}$ and direction of $\overrightarrow{A}-\overrightarrow{B}$ is $76.0°\text{above the -x axis}$.