Chapter 13.4, Problem 2CE

a.

To determine

To find the magnitude of the vector $\overrightarrow{OB}$ .

Answer to Problem 2CE

The magnitude of the vector $\overrightarrow{OB}$ is5.

Explanation of Solution

Given information: The vector is $\overrightarrow{OB}$ .

  

Calculation: The required magnitude is obtained as,

  $\overrightarrow{OB}=\sqrt{x^2+y^2}=\sqrt{4^2+{\left(-3\right)}^2}=\sqrt{25}=5$

Hence, the magnitude of the vector $\overrightarrow{OB}$ is 5.

b.

To determine

To find the magnitude of the vector $\overrightarrow{OD}$ .

Answer to Problem 2CE

The magnitude of the vector $\overrightarrow{OD}$ is $\sqrt{26}$ .

Explanation of Solution

Given information: The vector is $\overrightarrow{OD}$ .

  

Calculation: The required magnitude is obtained as,

  $\overrightarrow{OD}=\sqrt{x^2+y^2}=\sqrt{1^2+5^2}=\sqrt{26}$

Hence, the magnitude of the vector $\overrightarrow{OD}$ is $\sqrt{26}$ .

c.

To determine

To find the magnitude of the vector $\overrightarrow{DE}$ .

Answer to Problem 2CE

The magnitude of the vector $\overrightarrow{DE}$ is 5.

Explanation of Solution

Given information: The vector is $\overrightarrow{DE}$ .

  

Calculation: The required magnitude is obtained as,

  $\overrightarrow{DE}=\sqrt{x^2+y^2}=\sqrt{{\left(-5\right)}^2+0^2}=\sqrt{25}=5$

Hence, the magnitude of the vector $\overrightarrow{DE}$ is 5.

d.

To determine

To find the magnitude of the vector $\overrightarrow{EF}$ .

Answer to Problem 2CE

The magnitude of the vector $\overrightarrow{EF}$ is 2.

Explanation of Solution

Given information: The vector is $\overrightarrow{EF}$ .

  

Calculation: The required magnitude is obtained as,

  $\overrightarrow{EF}=\sqrt{x^2+y^2}=\sqrt{0^2+{\left(-2\right)}^2}=\sqrt{4}=2$

Hence, the magnitude of the vector $\overrightarrow{EF}$ is 2.

e.

To determine

To find the magnitude of the vector $\overrightarrow{BC}$ .

Answer to Problem 2CE

The magnitude of the vector $\overrightarrow{BC}$ is $\sqrt{26}$ .

Explanation of Solution

Given information: The vector is $\overrightarrow{BC}$ .

  

Calculation: The required magnitude is obtained as,

  $\overrightarrow{BC}=\sqrt{x^2+y^2}=\sqrt{1^2+5^2}=\sqrt{26}$

Hence, the magnitude of the vector $\overrightarrow{BC}$ is $\sqrt{26}$ .

f.

To determine

To find the magnitude of the vector $\overrightarrow{AG}$ .

Answer to Problem 2CE

The magnitude of the vector $\overrightarrow{AG}$ is 5.

Explanation of Solution

Given information: The vector is $\overrightarrow{AG}$ .

  

Calculation: The required magnitude is obtained as,

  $\overrightarrow{AG}=\sqrt{x^2+y^2}=\sqrt{4^2+{\left(-3\right)}^2}=\sqrt{25}=5$

Hence, the magnitude of the vector $\overrightarrow{AG}$ is 5.