A) \[{{\theta }_{1}}=\frac{\pi }{6},{{\theta }_{2}}=\frac{\pi }{3}\]
B) \[{{\theta }_{1}}=\frac{\pi }{3},{{\theta }_{2}}=\frac{\pi }{6}\]
C) \[{{\theta }_{1}}=\frac{\pi }{2},{{\theta }_{2}}=\frac{\pi }{3}\]
D) \[{{\theta }_{1}}=\frac{\pi }{3},{{\theta }_{2}}=\frac{\pi }{2}\]
Correct Answer: C
Solution :
Since, \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\frac{1}{2}\overrightarrow{b}\] \[\Rightarrow \] \[(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\,\overrightarrow{c}=\frac{1}{2}\overrightarrow{b}\] On comparing both sides, we get \[\overrightarrow{a}\cdot \overrightarrow{c}=\frac{1}{2}\]and \[\overrightarrow{a}\cdot \overrightarrow{b}=0\] Now, \[\overrightarrow{a}\cdot \,\overrightarrow{c}=\frac{1}{2}\] \[\Rightarrow \] \[\overrightarrow{a}\cdot \,\overrightarrow{c}\cos {{\theta }_{2}}=\frac{1}{2}\] \[\Rightarrow \]\[\cos {{\theta }_{2}}=\frac{1}{2}\] (\[\because \]\[\overrightarrow{a}\] and \[\overrightarrow{c}\] are unit vectors) \[\Rightarrow \] \[\cos {{\theta }_{2}}=\cos \frac{\pi }{3}\] \[\Rightarrow \] \[{{\theta }_{2}}=\frac{\pi }{3}\] and \[\overrightarrow{a}\cdot \overrightarrow{b}=0\] \[\Rightarrow \] \[|\overrightarrow{a}|\,|\overrightarrow{b}|cos{{\theta }_{1}}=0\] \[\Rightarrow \]\[\cos {{\theta }_{1}}=\cos \frac{\pi }{2}\] (\[\overrightarrow{a}\] and \[\overrightarrow{b}\] are unit vectors) \[\Rightarrow \] \[{{\theta }_{1}}=\frac{\pi }{2}\] Hence, \[{{\theta }_{1}}=\frac{\pi }{2}\] and \[{{\theta }_{2}}=\frac{\pi }{3}.\]
You need to login to perform this action.
You will be redirected in
3 sec
