A) \[1\]
B) \[1/4\]
C) \[0\]
D) \[-1/4\]
Correct Answer: D
Solution :
\[(\vec{a}-\lambda \vec{b}).(\vec{b}-2\vec{c})\times (\vec{c}+2\vec{a})=0\] \[\Rightarrow \] \[(\vec{a}-\lambda \vec{b}).\{\vec{b}\times \vec{c}+\vec{b}\times 2\vec{a}-2\vec{c}\times \vec{c}\] \[-4(\vec{c}\times \vec{a})\}=0\] \[\Rightarrow \]\[(\vec{a}-\lambda \vec{b}).\{\vec{b}\times \vec{c}+\vec{b}\times 2\vec{a}-4(\vec{c}\times \vec{a})\}=0\] \[\Rightarrow \]\[\vec{a}.(\vec{b}\times \vec{c})+\vec{a}.(\vec{b}\times 2\vec{a})-\vec{a}.4(\vec{c}\times \vec{a})\]\[-\lambda \vec{b}.(\vec{b}\times \vec{c})-\lambda \vec{b}.(\vec{b}\times 2\vec{a})\] \[+4\lambda \vec{b}.(\vec{c}\times \vec{a})=0\] \[\Rightarrow \] \[\vec{a}.(\vec{b}\times \vec{c})+4\lambda \vec{b}.(\vec{c}\times \vec{a})=0\] \[\Rightarrow \] \[\{\vec{a}.(\vec{b}\times \vec{c})\}(1+4\lambda )=0\]\[=\left( \frac{1}{\sqrt{26}} \right)\hat{i}+\left( \frac{4}{\sqrt{26}} \right)\hat{j}+\left( \frac{3}{\sqrt{26}} \right)\hat{k}\] \[\therefore \] \[\Rightarrow \] \[1+4\lambda =0\] \[[\because \,\,\vec{a}.(\vec{b}\times \vec{c})\ne 0]\] \[\Rightarrow \] \[\lambda =-\frac{1}{4}\]
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