Let a∧,b∧,c∧ be three pair-wise non collinear unit vectors, such thata∧×(b∧×c∧)=b∧3+c∧2 and b∧×(c∧×a∧)=−c∧2 then absolute value of [a∧b∧c∧] is

a∧×(b∧×c∧)=b∧3+c∧2⇒(a∧.c∧)b∧−(a∧.b∧)c∧=b∧3+c∧2⇒a∧.c∧=13;a∧.b∧=−12 Also b∧×(c∧×a∧)=−c∧2⇒(b∧.a∧)c∧−(b∧.c∧)a∧=−c∧2⇒b∧.a∧=−12;b∧.c∧=0∵[a∧b∧c∧]2=|a∧.a∧a∧.b∧a∧.c∧b∧.a∧b∧.b∧b∧.c∧c∧.a∧c∧.b∧c∧.c∧|⇒[a∧b∧c∧]2=|1−1213−12101301|=2336

Let a∧,b∧,c∧ be three pair-wise non collinear unit vectors, such thata∧×(b∧×c∧)=b∧3+c∧2 and b∧×(c∧×a∧)=−c∧2 then absolute value of [a∧b∧c∧] is

a∧×(b∧×c∧)=b∧3+c∧2⇒(a∧.c∧)b∧−(a∧.b∧)c∧=b∧3+c∧2⇒a∧.c∧=13;a∧.b∧=−12 Also b∧×(c∧×a∧)=−c∧2⇒(b∧.a∧)c∧−(b∧.c∧)a∧=−c∧2⇒b∧.a∧=−12;b∧.c∧=0∵[a∧b∧c∧]2=|a∧.a∧a∧.b∧a∧.c∧b∧.a∧b∧.b∧b∧.c∧c∧.a∧c∧.b∧c∧.c∧|⇒[a∧b∧c∧]2=|1−1213−12101301|=2336

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Let $\overset{\land}{a}, \overset{\land}{b}, \overset{\land}{c}$ be three pair-wise non collinear unit vectors, such that $\overset{\land}{a} \times (\overset{\land}{b} \times \overset{\land}{c}) = \frac{\overset{\land}{b}}{3} + \frac{\overset{\land}{c}}{2}$ and $\overset{\land}{b} \times (\overset{\land}{c} \times \overset{\land}{a}) = - \frac{\overset{\land}{c}}{2}$ then absolute value of $[\overset{\land}{a} \overset{\land}{b} \overset{\land}{c}]$ is

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$\frac{13}{36}$

$\frac{\sqrt{13}}{6}$

$\frac{\sqrt{23}}{6}$

$\frac{23}{36}$

answer is C.

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Detailed Solution

$\overset{\land}{a} \times (\overset{\land}{b} \times \overset{\land}{c}) = \frac{\overset{\land}{b}}{3} + \frac{\overset{\land}{c}}{2}$

$\Rightarrow (\overset{\land}{a} . \overset{\land}{c}) \overset{\land}{b} - (\overset{\land}{a} . \overset{\land}{b}) \overset{\land}{c} = \frac{\overset{\land}{b}}{3} + \frac{\overset{\land}{c}}{2}$

$\Rightarrow \overset{\land}{a} . \overset{\land}{c} = \frac{1}{3}; \overset{\land}{a} . \overset{\land}{b} = \frac{- 1}{2}$

Also $\overset{\land}{b} \times (\overset{\land}{c} \times \overset{\land}{a}) = \frac{- \overset{\land}{c}}{2}$

$\Rightarrow (\overset{\land}{b} . \overset{\land}{a}) \overset{\land}{c} - (\overset{\land}{b} . \overset{\land}{c}) \overset{\land}{a} = \frac{- \overset{\land}{c}}{2}$

$\Rightarrow \overset{\land}{b} . \overset{\land}{a} = \frac{- 1}{2}; \overset{\land}{b} . \overset{\land}{c} = 0$

$∵ {[\overset{\land}{a} \overset{\land}{b} \overset{\land}{c}]}^{2} = | \begin{matrix} \overset{\land}{a} . \overset{\land}{a} & \overset{\land}{a} . \overset{\land}{b} & \overset{\land}{a} . \overset{\land}{c} \\ \overset{\land}{b} . \overset{\land}{a} & \overset{\land}{b} . \overset{\land}{b} & \overset{\land}{b} . \overset{\land}{c} \\ \overset{\land}{c} . \overset{\land}{a} & \overset{\land}{c} . \overset{\land}{b} & \overset{\land}{c} . \overset{\land}{c} \end{matrix} |$

$\Rightarrow {[\overset{\land}{a} \overset{\land}{b} \overset{\land}{c}]}^{2} = | \begin{matrix} 1 & \frac{- 1}{2} & \frac{1}{3} \\ \frac{- 1}{2} & 1 & 0 \\ \frac{1}{3} & 0 & 1 \end{matrix} | = \frac{23}{36}$