question_answer

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