Question

The resultant of two forces $(A+B)$ and $(A-B)$ is a force $\sqrt{3A^2+B^2}.$ The angle between two given forces is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

Answer: (B)

Here, $\vec{P}=(\vec{A}+\vec{B}),\vec{Q}=(\vec{A}-\vec{B}),\vec{R}$
$=\sqrt{3A^2+B^2}$
Let $\theta$ be angle between the two given forces.
$\therefore \left(3A^2+B^2\right)=(A+B{)}^2+(A-B{)}^2+2(A+B)(A-B)\cos \theta$
or $3A^2+B^2=A^2+B^2+2AB+A^2+B^2-2AB+2\left(A^2-B^2\right)\cos \theta$
or $A^2-B^2=2\left(A^2-B^2\right)\cos \theta$
or $\cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}$ or
$\theta =\frac{\pi }{3}$