Let a, b and c be unit vectors such that a cd | vedclass

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(A) Given,$|a| = |b| = |c| = 1$ and $a \cdot b = 0 = a \cdot c$. The angle between $b$ and $c$ is $\frac{\pi}{3}$.We need to find $|a 	imes b - a 	i

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(A) Given,$|a| = |b| = |c| = 1$ and $a \cdot b = 0 = a \cdot c$. The angle between $b$ and $c$ is $\frac{\pi}{3}$.We need to find $|a 	imes b - a 	imes c|$.Using the distributive property of the cross product,we have $|a 	imes b - a 	imes c| = |a 	imes (b - c)|$.Since $a \cdot b = 0$ and $a \cdot c = 0$,we have $a \cdot (b - c) = 0$,which implies that $a$ is perpendicular to $(b - c)$.Thus,$|a 	imes (b - c)| = |a| |b - c| \sin \frac{\pi}{2} = |a| |b - c| (1) = |b - c|$.Now,calculate $|b - c|^2 = |b|^2 + |c|^2 - 2(b \cdot c) = 1 + 1 - 2(|b| |c| \cos \frac{\pi}{3}) = 2 - 2(1 \cdot 1 \cdot \frac{1}{2}) = 2 - 1 = 1$.Therefore,$|b - c| = 1$.Hence,$|a 	imes b - a 	imes c| = 1$.

Let $a, b$ and $c$ be unit vectors such that $a \cdot b = 0 = a \cdot c$ and the acute angle between $b$ and $c$ is $\frac{\pi}{3}$,then $|a \times b - a \times c|$ is equal to

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