Question

If $\overrightarrow{a},\text{ }\overrightarrow{b},\text{ }\overrightarrow{c}$ are non-coplanar and $(\overrightarrow{a}+\lambda \overrightarrow{b}).[(\overrightarrow{b}+3\overrightarrow{c})\times (\overrightarrow{c}\times 4\overrightarrow{a})]=0,$ then the value of $\lambda$ is equal to

• A. $0$
• B. $\frac{1}{12}$
• C. $\frac{5}{12}$
• D. $3$

Answer 1

Answer: (B)

$\frac{1}{12}$

Answer 2

Given, $(\overrightarrow{a}+\lambda \overrightarrow{b}).[(\overrightarrow{b}+3\overrightarrow{c})\times (\overrightarrow{c}-4\overrightarrow{a})]=0$
$\Rightarrow$ $(\overrightarrow{a}+\lambda \overrightarrow{b}).[\overrightarrow{b}\times \overrightarrow{c}-4\overrightarrow{b}\times \overrightarrow{a}-12\overrightarrow{c}\times \overrightarrow{a}]=0$
$\Rightarrow$ $[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]-0-0+0+0-12\lambda [\overrightarrow{b}\overrightarrow{c}a]=0$
$\Rightarrow$ $[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]=12\lambda [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]$
$\Rightarrow$ $\lambda =\frac{1}{12}$