A) \[\frac{1}{3}\]
B) \[\frac{\sqrt{2}}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{2\sqrt{2}}{3}\]
Correct Answer: D
Solution :
Since, \[\frac{1}{3}|b||c|a=(a\times b)\times c\] We know that \[(a\times b)\times c=(a.c)b-(b.c)a\] \[\therefore \] \[\frac{1}{3}|b||c|a=(a.c)b-(b.c)a\] On comparing the coefficients of a and b, we get \[\frac{1}{3}|b|c|=-b.c\]and\[a.c=0\] \[\Rightarrow \] \[\frac{1}{3}bc=-b\cos \theta \] \[\Rightarrow \] \[\cos \theta =-\frac{1}{3}\] \[\Rightarrow \] \[{{\cos }^{2}}\theta =\frac{1}{9}\] \[\Rightarrow \] \[1-{{\sin }^{2}}\theta =\frac{1}{9}\] \[\Rightarrow \] \[{{\sin }^{2}}\theta =1-\frac{1}{9}=\frac{8}{9}\] \[\Rightarrow \] \[\sin \theta =\frac{2\sqrt{2}}{3}\] \[(\because 0\le \theta \le \pi )\]
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