Chapter 3, Problem 39PQ

(a)

To determine

The vector $\overrightarrow{A}$ in component form.

Answer to Problem 39PQ

The vector $\overrightarrow{A}$ in component form is $\underset{\_}{\overrightarrow{A}=\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}}$.

Explanation of Solution

Write the expression to find the component form of any vector.

    $\overrightarrow{A}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

Here, $x$ is the $x$ component of the vector, $y$ is the $y$ component of the vector and $z$ is the $z$ component of the vector.

Conclusion:

Substitute $15.0\text{m}$ for $x$, $-6.00\text{m}$ for $y$ and $-3.00\text{m}$ for $z$ in the above equation.

    $\begin{array}{c}\overrightarrow{A}=\left(15.0\text{m}\right)\widehat{i}+\left(-6.00\text{m}\right)\widehat{j}+\left(-3.00\text{m}\right)\widehat{k}\\ =\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}\end{array}$

Therefore, the vector $\overrightarrow{A}$ in component form is $\underset{\_}{\overrightarrow{A}=\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}}$.

(b)

To determine

The vector $\overrightarrow{B}$ in component form having one-third the length of vector $\overrightarrow{A}$ but in the same direction of the vector $\overrightarrow{A}$.

Answer to Problem 39PQ

The vector $\overrightarrow{B}$ in component form having one-third the length of vector $\overrightarrow{A}$ but in the same direction of the vector $\overrightarrow{A}$ is $\underset{\_}{\overrightarrow{B}=\left(5.00\widehat{i}-2.00\widehat{j}-1.00\widehat{k}\right)\text{m}}$.

Explanation of Solution

Write the expression to find the vector $\overrightarrow{B}$.

    $\overrightarrow{B}=\frac{1}{3}\overrightarrow{A}$

Conclusion:

Substitute $\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}$ for $\overrightarrow{A}$ in the above equation.

    $\begin{array}{c}\overrightarrow{B}=\frac{1}{3}\left(\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}\right)\\ =\left(5.0\widehat{i}-2.00\widehat{j}-1.00\widehat{k}\right)\text{m}\end{array}$

Therefore, the vector $\overrightarrow{B}$ in component form having one-third the length of vector $\overrightarrow{A}$ but in the same direction of the vector $\overrightarrow{A}$ is $\underset{\_}{\overrightarrow{B}=\left(5.00\widehat{i}-2.00\widehat{j}-1.00\widehat{k}\right)\text{m}}$.

(c)

To determine

The vector $\overrightarrow{C}$ in component form having thrice the length of vector $\overrightarrow{A}$ but in the opposite direction of the vector $\overrightarrow{A}$.

Answer to Problem 39PQ

The vector $\overrightarrow{C}$ in component form having thrice the length of vector $\overrightarrow{A}$ but in the opposite direction of the vector $\overrightarrow{A}$ is $\underset{\_}{\overrightarrow{C}=\left(-45.00\widehat{i}+18.0\widehat{j}+9.00\widehat{k}\right)\text{m}}$.

Explanation of Solution

Write the expression to find the vector $\overrightarrow{C}$.

    $\overrightarrow{C}=-3\overrightarrow{A}$

Conclusion:

Substitute $\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}$ for $\overrightarrow{A}$ in the above equation.

    $\begin{array}{c}\overrightarrow{C}=-3\left(\left(15.0\widehat{i}-6.00\widehat{j}-3.00\widehat{k}\right)\text{m}\right)\\ =\left(-45.00\widehat{i}+18.0\widehat{j}+9.00\widehat{k}\right)\text{m}\end{array}$

Therefore, the vector $\overrightarrow{C}$ in component form having thrice the length of vector $\overrightarrow{A}$ but in the opposite direction of the vector $\overrightarrow{A}$ is $\underset{\_}{\overrightarrow{C}=\left(-45.00\widehat{i}+18.0\widehat{j}+9.00\widehat{k}\right)\text{m}}$.