Question

If B is 3km at an angle 30 degrees, determine the magnitude and direction of -2B

Answer 1

To determine the magnitude and direction of \(-2B\), we first need to analyze the vector \(B\).

Given:
- Magnitude of \(B = 3 \, \text{km}\)
- Direction of \(B\) = \(30^\circ\) from the positive x-axis

### Step 1: Find the components of \(B\)

Using trigonometry, we can find the components of vector \(B\):

- The x-component (\(B_x\)) is given by:
\[
B_x = |B| \cdot \cos(\theta) = 3 \cdot \cos(30^\circ) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.598 \, \text{km}
\]

- The y-component (\(B_y\)) is given by:
\[
B_y = |B| \cdot \sin(\theta) = 3 \cdot \sin(30^\circ) = 3 \cdot \frac{1}{2} = 1.5 \, \text{km}
\]

### Step 2: Find the components of \(-2B\)

Now, we can calculate the components of \(-2B\):

- The x-component of \(-2B\) is:
\[
-2B_x = -2 \cdot \frac{3\sqrt{3}}{2} = -3\sqrt{3} \approx -5.196 \, \text{km}
\]

- The y-component of \(-2B\) is:
\[
-2B_y = -2 \cdot 1.5 = -3 \, \text{km}
\]

### Step 3: Calculate the magnitude of \(-2B\)

The magnitude of \(-2B\) can be calculated using the Pythagorean theorem:

\[
| -2B | = \sqrt{(-3\sqrt{3})^2 + (-3)^2} = \sqrt{(27) + (9)} = \sqrt{36} = 6 \, \text{km}
\]

### Step 4: Find the direction of \(-2B\)

To find the direction, we calculate the angle \(\phi\) with respect to the positive x-axis:

\[
\tan(\phi) = \frac{B_y}{B_x} = \frac{-3}{-3\sqrt{3}} = \frac{1}{\sqrt{3}}
\]

Calculating \(\phi\):
\[
\phi = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ
\]

Since both components are negative, the vector \(-2B\) is located in the third quadrant. Therefore, we add \(180^\circ\) to \(30^\circ\):

\[
\text{Direction of } -2B = 30^\circ + 180^\circ = 210^\circ
\]

### Conclusion

The magnitude and direction of \(-2B\) are:
- Magnitude: **6 km**
- Direction: **210 degrees**