If a→×b→=c→,b→×c→=a→ and a,b,c be the moduli of the vectors a→,b→,c→, respectively, then

Since a→×b→=c→(b→×c→)×b→=c→⇒ (b→⋅b→)c→−(b→⋅c→)b→=c→⇒ (b→⋅b→)=1,c→⋅b→=0⇒ b2=1, i.e. b=1 Therefore, b→ is a unit vector. ∴ |c→|=|a→|, i.e., c=a.

If a→×b→=c→,b→×c→=a→ and a,b,c be the moduli of the vectors a→,b→,c→, respectively, then

Since a→×b→=c→(b→×c→)×b→=c→⇒ (b→⋅b→)c→−(b→⋅c→)b→=c→⇒ (b→⋅b→)=1,c→⋅b→=0⇒ b2=1, i.e. b=1 Therefore, b→ is a unit vector. ∴ |c→|=|a→|, i.e., c=a.

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$If \vec{a} \times \vec{b} = \vec{c}, \vec{b} \times \vec{c} = \vec{a} and a, b, c be the moduli of the vectors \vec{a}, \vec{b}, \vec{c}, respectively, then$

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$a = 1, b = 1$

$c = 1, a = 1$

$\vec{a} \cdot (\vec{b} \times \vec{c}) = - 1$

$b = 1, c = a$

answer is D.

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Detailed Solution

$Since \vec{a} \times \vec{b} = \vec{c}$

$\begin{array}{l} (\vec{b} \times \vec{c}) \times \vec{b} = \vec{c} \\ \Rightarrow (\vec{b} \cdot \vec{b}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{b} = \vec{c} \\ \Rightarrow (\vec{b} \cdot \vec{b}) = 1, \vec{c} \cdot \vec{b} = 0 \\ \Rightarrow b^{2} = 1, i.e. b = 1 \end{array}$