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.
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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: \[ B_x = B \cdot \cos(\theta) = 3 \cdot \cos(30^\circ) = 3 \cdot \frac{\sqrt{3}}{2} \approx 2.598 , \text{km} \] \[ B_y = B \cdot \sin(\theta) = 3 \cdot \sin(30^\circ) = 3 \cdot \frac{1}{2} = 1.5 , \text{km} \]
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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: \[ -B_x = -2.598 , \text{km} \] \[ -B_y = -1.5 , \text{km} \]
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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.
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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.