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