For three unit vectors a¯, b¯, c¯ no two of which are collinear satisfying a¯×(b¯×c¯)=12.b¯ Then angle between a¯ and b¯ is

(a¯⋅c¯)b¯−(a¯⋅b¯)⋅c¯=12⋅b¯⇒a¯⋅c¯−12b=(a¯⋅b¯)c¯∴a¯⋅c¯=12,a¯⋅b¯=0⇒θ=π2

For three unit vectors a¯, b¯, c¯ no two of which are collinear satisfying a¯×(b¯×c¯)=12.b¯ Then angle between a¯ and b¯ is

(a¯⋅c¯)b¯−(a¯⋅b¯)⋅c¯=12⋅b¯⇒a¯⋅c¯−12b=(a¯⋅b¯)c¯∴a¯⋅c¯=12,a¯⋅b¯=0⇒θ=π2

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For three unit vectors $\bar{a}, \bar{b}, \bar{c}$ no two of which are collinear satisfying $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{1}{2} . \bar{b}$ Then angle between $\bar{a}$ and $\bar{b}$ is

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$

answer is C.

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Detailed Solution

$\begin{array}{l} (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \cdot \bar{c} = \frac{1}{2} \cdot \bar{b} \\ \Rightarrow (\bar{a} \cdot \bar{c} - \frac{1}{2}) b = (\bar{a} \cdot \bar{b}) \bar{c} \\ ∴ \bar{a} \cdot \bar{c} = \frac{1}{2}, \bar{a} \cdot \bar{b} = 0 \Rightarrow θ = \frac{π}{2} \end{array}$