Chapter 1, Problem 32P

To determine

The value of $\text{A}+\text{B}$ and $\text{A}-\text{B}$ .

Explanation of Solution

Given:

The magnitude of vector A is $3.00\text{ units}$ and its direction is $30°$ above x -axis.

The magnitude of vector B is $2.00\text{ units}$ and its direction is $45°$ below − x -axis.

Formula Used:

The vector A is above positive x -axis; therefore, its component will be positive.

Write the expression for vector A.

  $\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}=\text{A}\left(\cos {\theta }_{\text{A}}\widehat{i}+\sin {\theta }_{\text{A}}\widehat{j}\right)$ ...... (I)

Here, $\cos {\theta }_{\text{A}}\widehat{i}$ is a horizontal component, $\sin {\theta }_{\text{A}}\widehat{j}$ is a vertical component and ${\theta }_{\text{A}}$ is the angle of vector A with the positive x -axis.

The vector B is below the negative x -axis; therefore, its component will be negative.

Write the expression for vector B.

  $\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}=-\text{B}\left(\cos {\theta }_{\text{B}}\widehat{i}+\sin {\theta }_{\text{B}}\widehat{j}\right)$ ...... (II)

Here, $\cos {\theta }_{\text{B}}\widehat{i}$ is a horizontal component, $\sin {\theta }_{\text{B}}\widehat{j}$ is a vertical component and ${\theta }_{\text{B}}$ is the angle of vector B with the negative y -axis.

Write the expression for the sum of two vectors.

  $\text{S}=\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}+\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}$ ...... (III)

Here, S is the sum of vectors.

Write the expression for the difference of two vectors.

  $\text{D}=\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}-\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}$ ...... (IV)

Calculation:

Substitute $3.00\text{ units}$ for A and $30°$ for ${\theta }_{\text{A}}$ in equation (I).

  $\begin{array}{c}\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}=\left(3.00\text{ units}\right)\left(\cos 30°\widehat{i}+\sin 30°\widehat{j}\right)\\ =2.6\widehat{i}+1.5\widehat{j}\end{array}$

Substitute $2.00\text{ units}$ for B and $45°$ for ${\theta }_{\text{B}}$ in equation (II).

  $\begin{array}{c}\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}=-\left(2.00\text{ units}\right)\left(\cos 45°\widehat{i}+\sin 45°\widehat{j}\right)\\ =-1.4\widehat{i}-1.4\widehat{j}\end{array}$

Substitute $2.6\widehat{i}+1.5\widehat{j}$ for $\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}$ and $-1.4\widehat{i}-1.4\widehat{j}$ for $\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}$ in equation (III).

  $\begin{array}{c}\text{S}=2.6\widehat{i}+1.5\widehat{j}+\left(-1.4\widehat{i}-1.4\widehat{j}\right)\\ =1.2\widehat{i}+0.1\widehat{j}\end{array}$

Substitute $2.6\widehat{i}+1.5\widehat{j}$ for $\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}$ and $-1.4\overrightarrow{i}-1.4\overrightarrow{j}$ for $\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}$ in equation (III).

  $\begin{array}{c}\text{D}=2.6\widehat{i}+1.5\widehat{j}-\left(-1.4\widehat{i}-1.4\widehat{j}\right)\\ =4.0\widehat{i}+2.9\widehat{j}\end{array}$

Conclusion:

Thus, the value of $\text{A}+\text{B}$ is $1.2\widehat{i}+0.1\widehat{j}$ and the value of $\text{A}-\text{B}$ is $4.0\widehat{i}+2.9\widehat{j}$ .