A) \[1/4\]
B) \[1/3\]
C) \[3/4\]
D) \[1/2\]
Correct Answer: C
Solution :
[c] \[\vec{u}\times (\vec{u}\times \vec{w})+3\vec{v}=0\] \[\Rightarrow \,\,\,\,(\vec{u}.\vec{w})\vec{u}-(\vec{u}.\vec{u})w=-3\vec{v}\] \[\Rightarrow \,\,\,\,\left( 2\sqrt{3}\cos \theta \right)\vec{u}-3\vec{w}=-3\vec{v}\] Squaring both sides, we get \[{{\left| {\vec{u}} \right|}^{2}}{{\left( 2\sqrt{3}\cos \theta \right)}^{2}}+9{{\left| {\vec{w}} \right|}^{2}}-12\sqrt{3}\cos \theta \left| {\vec{u}} \right|\left| {\vec{w}} \right|\cos \theta =9{{\left| {\vec{v}} \right|}^{2}}\]\[\Rightarrow \,\,\,\,{{\cos }^{2}}\theta =\frac{3}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
