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(C) Let $\vec{u} = \sqrt{3}(\vec{a} 	imes \vec{b})$ and $\vec{v} = \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.We know that $\vec{v} = \vec{a} 	imes (

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$\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$

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(C) Let $\vec{u} = \sqrt{3}(\vec{a} 	imes \vec{b})$ and $\vec{v} = \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.We know that $\vec{v} = \vec{a} 	imes (\vec{b} 	imes \vec{a})$.Since $\vec{a} 	imes \vec{b}$ is perpendicular to $\vec{a}$ and $\vec{b}$,and $\vec{a} 	imes (\vec{b} 	imes \vec{a})$ is also perpendicular to $\vec{a}$,we observe that $\vec{u} \perp \vec{v}$.Thus,the triangle is a right-angled triangle with one angle equal to $\frac{\pi}{2}$.Now,$|\vec{u}| = \sqrt{3}|\vec{a} 	imes \vec{b}| = \sqrt{3}|\vec{a}||\vec{b}|\sin 	heta = \sqrt{3}|\vec{b}|\sin 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.Also,$|\vec{v}| = |\vec{a} 	imes (\vec{b} 	imes \vec{a})| = |\vec{a}| |\vec{b} 	imes \vec{a}| \sin(90^{\circ}) = |\vec{b}|\sin 	heta$.In the right-angled triangle,let $\alpha$ be the angle between the sides $\vec{u}$ and $\vec{v}$.Then $	an \alpha = \frac{|\vec{u}|}{|\vec{v}|} = \frac{\sqrt{3}|\vec{b}|\sin 	heta}{|\vec{b}|\sin 	heta} = \sqrt{3}$.Therefore,$\alpha = \frac{\pi}{3}$.The third angle is $\pi - \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.Thus,the angles are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$.

Let $\vec{a}$ be a unit vector and $\vec{b}$ be a nonzero vector not parallel to $\vec{a}$. The angles of the triangle,two of whose sides are represented by $\sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$,are

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