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 \[{{R}^{2}}={{C}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] \[\Rightarrow \] \[{{C}^{2}}={{C}^{2}}+2AB\cos \theta \] \[\Rightarrow \] \[\cos \theta =0\] \[(\because AB\ne 0)\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\]You need to login to perform this action.
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