If a, b, c are three unit vectors such that a | vedclass

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(B) Given that $a 	imes (b 	imes c) = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$. Using the vector triple product formula,$a 	imes (b 	imes c) = (a \cd

Vedclass · Accepted Answer

$120^{\circ}, 30^{\circ}$

Vedclass · Answer

(B) Given that $a 	imes (b 	imes c) = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$. Using the vector triple product formula,$a 	imes (b 	imes c) = (a \cdot c) b - (a \cdot b) c$. Comparing this with the given equation: $(a \cdot c) b - (a \cdot b) c = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$. Since $a, b, c$ are unit vectors,let $\alpha$ be the angle between $a$ and $c$,and $\beta$ be the angle between $a$ and $b$. Then $a \cdot c = \cos \alpha$ and $a \cdot b = \cos \beta$. Substituting these into the equation: $(\cos \alpha) b - (\cos \beta) c = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$. Comparing the coefficients of $b$ and $c$: $\cos \alpha = \frac{\sqrt{3}}{2} \implies \alpha = 30^{\circ}$. $-\cos \beta = \frac{1}{2} \implies \cos \beta = -\frac{1}{2} \implies \beta = 120^{\circ}$. Thus,the angle between $a$ and $b$ is $120^{\circ}$ and the angle between $a$ and $c$ is $30^{\circ}$. Therefore,option $(B)$ is correct.

If $a, b, c$ are three unit vectors such that $a \times (b \times c) = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$,then the angles between $a, b$ and $a, c$ respectively are

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