Chapter 3, Problem 32QAP

To determine

(a)

The components of the vector $\overrightarrow{A}+\overrightarrow{B}$

Answer to Problem 32QAP

The x component of the vector $\overrightarrow{A}+\overrightarrow{B}$ is 13 units and its y component is 6 units.

Explanation of Solution

Given:

The components of the vectors $\overrightarrow{A},\overrightarrow{B}\text{ and }\overrightarrow{C}$

  $\begin{array}{l}{A}_x=6\\ A_y=9\\ B_x=7\\ B_y=-3\\ C_x=0\\ C_y=-6\end{array}$

Formula used:

If $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{P}$, then the x component of $\overrightarrow{P}$ is given by,

  $P_x=A_x+B_x$

And the y component of $\overrightarrow{P}$ is given by,

  $P_y=A_y+B_y$

Calculation:

Substitute the given values of the variables in the formula and calculate the x component.

  $\begin{array}{c}{P}_x=A_x+B_x\\ =6+7\\ =13\end{array}$

Substitute the given values of the variables in the formula and calculate the y component.

  $\begin{array}{c}{P}_y=A_y+B_y\\ =9-3\\ =6\end{array}$

Conclusion:

The x component of the vector $\overrightarrow{A}+\overrightarrow{B}$ is 13 units and its y component is 6 units.

To determine

(b)

The components of the vector $\overrightarrow{A}-2\overrightarrow{C}$.

Answer to Problem 32QAP

The x component of the vector $\overrightarrow{A}-2\overrightarrow{C}$ is 6 units and the y component is 21 units.

Explanation of Solution

Given:

The components of the vectors $\overrightarrow{A},\overrightarrow{B}\text{ and }\overrightarrow{C}$

  $\begin{array}{l}{A}_x=6\\ A_y=9\\ B_x=7\\ B_y=-3\\ C_x=0\\ C_y=-6\end{array}$

Formula used:

If $\overrightarrow{A}-2\overrightarrow{C}=\overrightarrow{Q}$, then the x component of the vector $\overrightarrow{Q}$ is given by,

  $Q_x=A_x-2C_x$

The y component is given by,

  $Q_y=A_y-2C_y$

Calculation:

Substitute the given values of the variables in the formula and calculate the x component.

  $\begin{array}{c}{Q}_x=A_x-2C_x\\ =6-2\times 0\\ =6\end{array}$

Substitute the given values of the variables in the formula and calculate the y component.

  $\begin{array}{c}{Q}_y=A_y-2C_y\\ =9-2\times \left(-6\right)\\ =21\end{array}$

Conclusion:

The x component of the vector $\overrightarrow{A}-2\overrightarrow{C}$ is 6 units and the y component is 21 units.

To determine

(c)

The components of the vector $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$.

Answer to Problem 32QAP

The x component of the vector $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$ is 13 units and the y component is 12 units.

Explanation of Solution

Given:

The components of the vectors $\overrightarrow{A},\overrightarrow{B}\text{ and }\overrightarrow{C}$

  $\begin{array}{l}{A}_x=6\\ A_y=9\\ B_x=7\\ B_y=-3\\ C_x=0\\ C_y=-6\end{array}$

Formula used:

If $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}=\overrightarrow{R}$, then, the x component of the vector $\overrightarrow{R}$ is given by,

  $R_x=A_x+B_x-C_x$

The y component is given by,

  $R_y=A_y+B_y-C_y$

Calculation:

Substitute the given values of the variables in the formula and calculate the x component.

$\begin{array}{c}{R}_x=A_x+B_x-C_x\\ =6+7-0\\ =13\end{array}$

Substitute the given values of the variables in the formula and calculate the y component.

$\begin{array}{c}{R}_y=A_y+B_y-C_y\\ =9+\left(-3\right)-\left(-6\right)\\ =12\end{array}$

Conclusion:

The x component of the vector $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$ is 13 units and the y component is 12 units.

To determine

(d)

The components of the vector $\overrightarrow{A}+\frac{1}{2}\overrightarrow{B}-3\overrightarrow{C}$.

Answer to Problem 32QAP

The x component of the vector $\overrightarrow{A}+\frac{1}{2}\overrightarrow{B}-3\overrightarrow{C}$ is 9.5 units and the y component is 25.5 units.

Explanation of Solution

Given:

The components of the vectors $\overrightarrow{A},\overrightarrow{B}\text{ and }\overrightarrow{C}$

  $\begin{array}{l}{A}_x=6\\ A_y=9\\ B_x=7\\ B_y=-3\\ C_x=0\\ C_y=-6\end{array}$

Formula used:

If $\overrightarrow{A}+\frac{1}{2}\overrightarrow{B}-3\overrightarrow{C}=\overrightarrow{S}$, then its x component is given by,

  $S_x=A_x+\frac{1}{2}{B}_x-3C_x$

The y component is given by,

  $S_y=A_y+\frac{1}{2}{B}_y-3C_y$

Calculation:

Substitute the given values of the variables in the formula and calculate the x component of the vector $\overrightarrow{A}+\frac{1}{2}\overrightarrow{B}-3\overrightarrow{C}$.

  $\begin{array}{c}{S}_x=A_x+\frac{1}{2}{B}_x-3C_x\\ =6+\frac{1}{2}\times 7-3\times 0\\ =9.5\end{array}$

Substitute the given values of the variables in the formula and calculate the y component of the vector $\overrightarrow{A}+\frac{1}{2}\overrightarrow{B}-3\overrightarrow{C}$.

  $\begin{array}{c}{S}_y=A_y+\frac{1}{2}{B}_y-3C_y\\ =9+\frac{1}{2}\left(-3\right)-3\left(-6\right)\\ =25.5\end{array}$

Conclusion:

The x component of the vector $\overrightarrow{A}+\frac{1}{2}\overrightarrow{B}-3\overrightarrow{C}$ is 9.5 units and the y component is 25.5 units.