The angle between vector \( (\vec{A}) \) and \( (\vec{A} - \vec{B}) \) is:
The angle between vector \( (\vec{A}) \) and \( (\vec{A} - \vec{B}) \) is:
Show Hint
- \( \tan^{-1} \left( \frac{A}{0.7 B} \right) \)
- \( \tan^{-1} \left( \frac{\sqrt{3} B}{2A - B} \right) \)
- \( \tan^{-1} \left( \frac{B \cos \theta}{A - B \sin \theta} \right) \)
- \( \tan^{-1} \left( \frac{B/2}{A - B\sqrt{3}/2} \right) \)
\textbf{Correct Answer:} (B) \( \tan^{-1} \left( \frac{\sqrt{3} B}{2A - B} \right) \)
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are required to find the angle, say \(\beta\), between the vector \( \vec{A} \) and the vector \( \vec{A} - \vec{B} \). From the diagram, the angle between vectors \( \vec{A} \) and \( \vec{B} \) is \(120^\circ\). Important Clarification:
When dealing with vector subtraction, the vector \( \vec{A} - \vec{B} \) is equivalent to adding vector \( \vec{A} \) with vector \( -\vec{B} \). Hence, the relevant angle for component resolution is the angle between \( \vec{A} \) and \( -\vec{B} \), not between \( \vec{A} \) and \( \vec{B} \). Since the angle between \( \vec{A} \) and \( \vec{B} \) is \(120^\circ\), the angle between \( \vec{A} \) and \( -\vec{B} \) becomes: \[ 180^\circ - 120^\circ = 60^\circ \] Step 2: Choosing a Coordinate System:
Let vector \( \vec{A} \) be along the positive x-axis. \[ \vec{A} = A\hat{i} \] The vector \( -\vec{B} \) makes an angle of \(60^\circ\) with \( \vec{A} \). Step 3: Resolving Vectors into Components:
The components of vector \( -\vec{B} \) are: \[ -\vec{B} = B\cos 60^\circ\,\hat{i} + B\sin 60^\circ\,\hat{j} \] \[ -\vec{B} = \frac{B}{2}\hat{i} + \frac{\sqrt{3}B}{2}\hat{j} \] Now, the resultant vector: \[ \vec{R} = \vec{A} - \vec{B} = \vec{A} + (-\vec{B}) \] \[ \vec{R} = \left(A - \frac{B}{2}\right)\hat{i} - \frac{\sqrt{3}B}{2}\hat{j} \] Step 4: Calculating the Angle Between \( \vec{A} \) and \( \vec{R} \):
The angle \(\beta\) that \( \vec{R} \) makes with the x-axis (direction of \( \vec{A} \)) is: \[ \tan \beta = \frac{|R_y|}{|R_x|} \] \[ \tan \beta = \frac{\frac{\sqrt{3}B}{2}}{\frac{2A - B}{2}} \] \[ \tan \beta = \frac{\sqrt{3}B}{2A - B} \] Step 5: Final Answer:
\[ \beta = \tan^{-1}\left( \frac{\sqrt{3} B}{2A - B} \right) \] This matches option (B).
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