Question

The angle between \(\vec{A}\) and the resultant of \(2\vec{A}+3\vec{B}\) and \(4\vec{A}-3\vec{B}\) is

• A. \(90^\circ\)
• B. \(\tan^{-1}\left(\dfrac{A}{B}\right)\)
• C. \(\tan^{-1}\left(\dfrac{B}{A}\right)\)
• D. \(\tan^{-1}\left(\dfrac{A-B}{A+B}\right)\)
• E. \(0^\circ\)

Hint:

First add the vectors carefully. If the \(\vec{B}\) terms cancel, the resultant may become parallel to \(\vec{A}\).

Answer 1

Answer: (E)

The resultant of the two vectors is: \[ (2\vec{A}+3\vec{B})+(4\vec{A}-3\vec{B})=6\vec{A} \] So the resultant is in the same direction as \(\vec{A}\). Hence, the angle between \(\vec{A}\) and \(6\vec{A}\) is: \[ \boxed{0^\circ} \] Thus, \[ \boxed{(E)\ 0^\circ} \]