Question

The angle between vector \( (\vec{A}) \) and \( (\vec{A} - \vec{B}) \) is: 

 

• A. \( \tan^{-1} \left( \frac{A}{0.7 B} \right) \)
• B. \( \tan^{-1} \left( \frac{\sqrt{3} B}{2A - B} \right) \)
• C. \( \tan^{-1} \left( \frac{B \cos \theta}{A - B \sin \theta} \right) \)
• D. \( \tan^{-1} \left( \frac{B/2}{A - B\sqrt{3}/2} \right) \)
  \textbf{Correct Answer:} (B) \( \tan^{-1} \left( \frac{\sqrt{3} B}{2A - B} \right) \)

Hint:

While performing vector subtraction, always remember that \( \vec{A} - \vec{B} = \vec{A} + (-\vec{B}) \). Hence, the angle used in component resolution must be the angle between \( \vec{A} \) and \( -\vec{B} \), not \( \vec{B} \).

Answer 1

The Correct Option is B

To find the angle between the vectors (\vec{A}) and (\vec{A} - \vec{B}), we use the dot product formula:

\vec{A} \cdot (\vec{A} - \vec{B}) = |\vec{A}||\vec{A} - \vec{B}|\cos \theta

Given the vectors, this expands to:

\vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} = |\vec{A}||\vec{A} - \vec{B}|\cos \theta

Hence,

A^2 - AB \cos \theta' = A \sqrt{A^2 + B^2 - 2AB \cos \theta'} \cos \theta

Given that the angle \theta' is 120° as seen in the diagram:

\cos 120^\circ = -\frac{1}{2}

Substitute \cos 120^\circ in the equation:

A^2 + \frac{AB}{2} = A \sqrt{A^2 + B^2 + AB} \cos \theta

Rearrange and solve for \cos \theta:

\cos \theta = \frac{A^2 + \frac{AB}{2}}{A \sqrt{A^2 + B^2 + AB}}

Using the identity \tan \theta = \frac{\sin \theta}{\cos \theta}, calculate \tan \theta as follows:

\tan \theta = \frac{\sqrt{3} B}{2A - B}

The option that matches this result is:

(B) \tan^{-1} \left( \frac{\sqrt{3} B}{2A - B} \right)

The correct answer is, therefore, option (B).