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(D) The cross product of two non-zero vectors $u$ and $v$ is given by $u 	imes v = |u||v| \sin 	heta \hat{n}$,where $	heta$ is the angle between th

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$\leq |u||v|$

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(D) The cross product of two non-zero vectors $u$ and $v$ is given by $u 	imes v = |u||v| \sin 	heta \hat{n}$,where $	heta$ is the angle between the vectors and $\hat{n}$ is a unit vector perpendicular to both $u$ and $v$. Taking the magnitude on both sides,we get $|u 	imes v| = |u||v| |\sin 	heta|$. Since the range of the sine function is $[-1, 1]$,the absolute value $|\sin 	heta|$ satisfies $0 \leq |\sin 	heta| \leq 1$. Therefore,$|u 	imes v| = |u||v| |\sin 	heta| \leq |u||v|$. Thus,the magnitude of the cross product is always less than or equal to the product of the magnitudes of the individual vectors.

Let $u$ and $v$ be two non-zero vectors. Then the magnitude of the cross product $u \times v$ is always

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