A) 2
B) 4
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
\[\vec{a}\times \vec{b}\,=2\vec{a}\times \vec{c}\Rightarrow \vec{a}\times (\vec{b}-2\vec{c})=0\] \[\Rightarrow \,\vec{b}-2\vec{c}=\lambda \vec{a}\] But \[{{(\vec{b}-2\vec{c})}^{2}}\,={{\vec{b}}^{2}}+4{{\vec{c}}^{2}}-4\vec{b}.\vec{c}\] \[=16\,+4-4\times 4\,\times \,\frac{1}{4}=16\] Hence, \[{{\lambda }^{2}}{{a}^{2}}=16\Rightarrow \lambda =4\].You need to login to perform this action.
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