question_answer

Two vectors \[\vec{A}\] and \[\vec{B}\] are such that \[\vec{A}+\vec{B}=\vec{C}\]and \[{{A}^{2}}+{{B}^{2}}={{C}^{2}}\]. If \[\theta \] is the angle between \[\vec{A}\] and \[\vec{B}\] then correct statement is :

A)  \[\theta =\pi \]

B)  \[\theta =\frac{2\pi }{3}\]

C)  \[\theta =0\]

D)  \[\theta =\frac{\pi }{2}\]

Correct Answer: D

Solution :

If two vectors \[\vec{A}\] and \[\vec{B}\] inclined at angle \[\theta \] are taken then their resultant is \[R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] Given,   \[\vec{A}+\vec{B}=\vec{C}\] and        \[{{A}^{2}}+{{B}^{2}}={{C}^{2}}\] Squaring Eq. (i), we have                 \[{{R}^{2}}={{C}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] \[\Rightarrow \] \[{{C}^{2}}={{C}^{2}}+2AB\cos \theta \] \[\Rightarrow \] \[\cos \theta =0\] \[(\because AB\ne 0)\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\]