A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{\pi }{3}\]
Correct Answer: D
Solution :
Since, \[\vec{a}+\vec{b}+\vec{c}=\vec{0}\] \[\Rightarrow \]\[\vec{a}+\vec{b}=-\vec{c}\] \[\Rightarrow \]\[{{(\vec{a}+\vec{b})}^{2}}={{(-\vec{c})}^{2}}\] \[\Rightarrow \]\[|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+2|\vec{a}||\vec{b}|\cos \theta =|\vec{c}{{|}^{2}}\] \[\Rightarrow \] \[9+16+2.3.4\cos \theta =37\] \[\Rightarrow \] \[24\,\cos \theta =37-25\] \[\Rightarrow \] \[\cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]
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