If a→,b→ and c→ be four non-zero vectors such that [a→b→c→]=1,c→=λa→×b→ angle between a→ and b→ is 2π/3,|a→|=2|b→|=3 and |c→|=13 the angle between a→ and b→ is

c→=λ(a→×b→) or c→⋅c→=λ(a→×b→)⋅c→⇒ 13=λ Also |c→|2=λ2|a→×b→|2⇒ 13=19a2b2sin2⁡θ=19×2×3sin2⁡θ or sin2⁡θ=12 or θ=π4

If a→,b→ and c→ be four non-zero vectors such that [a→b→c→]=1,c→=λa→×b→ angle between a→ and b→ is 2π/3,|a→|=2|b→|=3 and |c→|=13 the angle between a→ and b→ is

c→=λ(a→×b→) or c→⋅c→=λ(a→×b→)⋅c→⇒ 13=λ Also |c→|2=λ2|a→×b→|2⇒ 13=19a2b2sin2⁡θ=19×2×3sin2⁡θ or sin2⁡θ=12 or θ=π4

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If $\vec{a}, \vec{b} and \vec{c}$ be four non-zero vectors such that $[\vec{a} \vec{b} \vec{c}] = 1, \vec{c} = λ \vec{a} \times \vec{b}$ angle between $\vec{a} and \vec{b} is 2 π / 3, | \vec{a} | = \sqrt{2}$
$| \vec{b} | = \sqrt{3} and | \vec{c} | = \frac{1}{\sqrt{3}}$ the angle between $\vec{a} and \vec{b}$ is

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$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} \vec{c} = λ (\vec{a} \times \vec{b}) \\ or \vec{c} \cdot \vec{c} = λ (\vec{a} \times \vec{b}) \cdot \vec{c} \\ \Rightarrow \frac{1}{3} = λ \\ Also | \vec{c} |^{2} = λ^{2} | \vec{a} \times \vec{b} |^{2} \\ \Rightarrow \frac{1}{3} = \frac{1}{9} (a^{2} b^{2} \sin^{2} θ) = \frac{1}{9} \times 2 \times 3 \sin^{2} θ \\ or \sin^{2} θ = \frac{1}{2} \\ or θ = \frac{π}{4} \end{array}$