Question

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

• A. $\pi /4$
• B. $\pi /3$
• C. $\pi /2$
• D. $\pi$

Answer 1

Answer: (B)

$\pi /3$

$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}\angle \overrightarrow{C}\overrightarrow{D}=\theta \angle \overrightarrow{A}\overrightarrow{B}=\phi R=\sqrt{C^2+D^2+2CD\cos \theta }$

$=\sqrt{(A^2+B^2+2AB\cos \phi )+(A^2+B^2-2AB\cos \phi )+2(A^2-B^2)\cos \theta }=\sqrt{2A^2+2B^2+2(A^2-B^2)\cos \theta }$

$R=\sqrt{3A^2+B^2}\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$