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(C) Given that $a, b$ and $c$ are mutually perpendicular vectors such that $a 	imes b = c$ and $b 	imes c = a$.Since the vectors are mutually perpen

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$1$ only

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(C) Given that $a, b$ and $c$ are mutually perpendicular vectors such that $a 	imes b = c$ and $b 	imes c = a$.Since the vectors are mutually perpendicular,we have $a \cdot b = 0$,$b \cdot c = 0$,and $c \cdot a = 0$.Also,the magnitude of the cross product is given by $|a 	imes b| = |a||b| \sin(90^\circ) = |a||b| = |c|$.Similarly,$|b 	imes c| = |b||c| = |a|$.Substituting $|a| = |b||c|$ into the first equation: $|b||c| \cdot |b| = |c|$.Since $c$ is a vector,$|c| 
eq 0$,so we can divide by $|c|$ to get $|b|^2 = 1$.Since the magnitude $|b|$ must be non-negative,$|b| = 1$.

If $a, b$ and $c$ are three vectors such that $a \times b = c$,$b \times c = a$ and $a, b, c$ are mutually perpendicular to each other,then $|b|$ is equal to

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