A) \[0\]
B) \[\frac{1}{12}\]
C) \[\frac{5}{12}\]
D) \[3\]
E) \[\frac{7}{12}\]
Correct Answer: B
Solution :
Given, \[(\overrightarrow{a}+\lambda \overrightarrow{b}).[(\overrightarrow{b}+3\overrightarrow{c})\times (\overrightarrow{c}-4\overrightarrow{a})]=0\] \[\Rightarrow \] \[(\overrightarrow{a}+\lambda \overrightarrow{b}).[\overrightarrow{b}\times \overrightarrow{c}-4\overrightarrow{b}\times \overrightarrow{a}-12\overrightarrow{c}\times \overrightarrow{a}]=0\] \[\Rightarrow \] \[[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]-0-0+0+0-12\lambda [\overrightarrow{b}\overrightarrow{c}a]=0\] \[\Rightarrow \] \[[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]=12\lambda [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]\] \[\Rightarrow \] \[\lambda =\frac{1}{12}\]
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