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

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(C) Given that $a, b, c$ are unit vectors, so $|a|=|b|=|c|=1$. Given $a+b+c=0$. The angle between $a$ and $b$ is $\frac{\pi}{3}$. Taking the cross pro

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$\frac{3 \sqrt{3}}{2}$

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(C) Given that $a, b, c$ are unit vectors, so $|a|=|b|=|c|=1$. Given $a+b+c=0$. The angle between $a$ and $b$ is $\frac{\pi}{3}$. Taking the cross product of $a+b+c=0$ with $a$: $a 	imes (a+b+c) = a 	imes 0$ $a 	imes a + a 	imes b + a 	imes c = 0$ Since $a 	imes a = 0$, we have $a 	imes b = c 	imes a$. Taking the magnitude, $|a 	imes b| = |c 	imes a|$. Similarly, by taking the cross product with $b$, we get $|a 	imes b| = |b 	imes c|$. Thus, $|a 	imes b| = |b 	imes c| = |c 	imes a|$. Therefore, $|a 	imes b| + |b 	imes c| + |c 	imes a| = 3|a 	imes b|$. Using the formula $|a 	imes b| = |a||b| \sin(	heta)$, where $	heta = \frac{\pi}{3}$: $|a 	imes b| = 1 	imes 1 	imes \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$. So, $3|a 	imes b| = 3 	imes \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$.

If $a, b$ and $c$ are unit vectors such that $a+b+c=0$ and the angle between $a$ and $b$ is $\frac{\pi}{3}$, then $|a \times b|+|b \times c|+|c \times a|=$

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