b¯,c¯ are unit vectors and |a¯|=7. a¯×(b¯×c¯)+b¯×(c¯×a¯)=12a¯ then the angle between a¯ and c¯is

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

b¯,c¯ are unit vectors and |a¯|=7. a¯×(b¯×c¯)+b¯×(c¯×a¯)=12a¯ then the angle between a¯ and c¯is

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

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$\bar{b}, \bar{c}$ are unit vectors and $| \bar{a} | = 7 . \bar{a} \times (\bar{b} \times \bar{c}) + \bar{b} \times (\bar{c} \times \bar{a}) = \frac{1}{2} \bar{a}$ then the angle between $\bar{a} and \bar{c}$ is

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{2}$

answer is C.

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$\Rightarrow (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} + (\bar{a} \cdot \bar{b}) \bar{c} - (\bar{b} \cdot \bar{c}) \bar{a} = \frac{1}{2} \bar{a}$

$\Rightarrow (\bar{a} \cdot \bar{c}) \bar{b} - ((\bar{b} \cdot \bar{c}) + \frac{1}{2}) \bar{a} = \bar{0}$

$\Rightarrow \bar{a} \cdot \bar{c} = 0, \bar{b} \cdot \bar{c} = - \frac{1}{2} \Rightarrow (\bar{a}, \bar{c}) = \frac{π}{2}$

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b¯,c¯ are unit vectors and |a¯|=7. a¯×(b¯×c¯)+b¯×(c¯×a¯)=12a¯ then the angle between a¯ and c¯is

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$\bar{b}, \bar{c}$ are unit vectors and $| \bar{a} | = 7 . \bar{a} \times (\bar{b} \times \bar{c}) + \bar{b} \times (\bar{c} \times \bar{a}) = \frac{1}{2} \bar{a}$ then the angle between $\bar{a} and \bar{c}$ is