Question

If B is 3km at an angle of 30 degrees. Determine the magnitude and direction of -B

Answer 1

To determine the magnitude and direction of \(-B\), we first need to analyze the vector \(B\). Given that vector \(B\) is 3 km at an angle of 30 degrees, we can find its components in terms of the x (horizontal) and y (vertical) axes using trigonometric functions.

1. **Calculate the components of vector \(B\)**:
- **Magnitude of \(B\)**: 3 km
- **Angle**: 30 degrees
- The x-component (\(B_x\)) and y-component (\(B_y\)) can be calculated as follows:
\[
B_x = B \cdot \cos(\theta) = 3 \cdot \cos(30^\circ) = 3 \cdot \frac{\sqrt{3}}{2} \approx 2.598 \, \text{km}
\]
\[
B_y = B \cdot \sin(\theta) = 3 \cdot \sin(30^\circ) = 3 \cdot \frac{1}{2} = 1.5 \, \text{km}
\]

2. **Finding \(-B\)**:
- The magnitude of \(-B\) is the same as that of \(B\), which is 3 km.
- The direction of \(-B\) is the opposite of vector \(B\). Therefore, we will negate both the x and y components:
\[
-B_x = -2.598 \, \text{km}
\]
\[
-B_y = -1.5 \, \text{km}
\]

3. **Determine the angle of \(-B\)**:
- To find the angle of \(-B\), we use the arctangent of the components:
\[
\theta' = \tan^{-1}\left(\frac{-B_y}{-B_x}\right) = \tan^{-1}\left(\frac{1.5}{2.598}\right)
\]
- This computation gives us the angle in the positive quadrant. However, since we are interested in the direction of \(-B\), it will be in the third quadrant because both components are negative.

4. **Calculate the angle in the third quadrant**:
\[
\theta' \approx 30^\circ \quad \text{(in the positive x-y coordinate system)}
\]
So, in the third quadrant, we find the angle as:
\[
180^\circ + 30^\circ = 210^\circ
\]

**Final Result:**
- The magnitude of \(-B\) is **3 km**.
- The direction of \(-B\) is **210 degrees** from the positive x-axis.