If a and b are two vectors,then (a times b)^2 | vedclass

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(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a 	imes b| = |a| |b| \sin 	heta$,where $	heta$ is the

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$\left| \begin{matrix} a \cdot a & a \cdot b \ b \cdot a & b \cdot b \end{matrix} ight|$

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(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a 	imes b| = |a| |b| \sin 	heta$,where $	heta$ is the angle between them.Therefore,$(a 	imes b)^2 = |a|^2 |b|^2 \sin^2 	heta$.Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get $(a 	imes b)^2 = |a|^2 |b|^2 (1 - \cos^2 	heta) = |a|^2 |b|^2 - |a|^2 |b|^2 \cos^2 	heta$.Since $a \cdot b = |a| |b| \cos 	heta$,we have $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 	heta$.Thus,$(a 	imes b)^2 = |a|^2 |b|^2 - (a \cdot b)^2$.Note that $|a|^2 = a \cdot a$ and $|b|^2 = b \cdot b$.So,$(a 	imes b)^2 = (a \cdot a)(b \cdot b) - (a \cdot b)(b \cdot a)$.This expression is equivalent to the determinant of the Gram matrix:$\left| \begin{matrix} a \cdot a & a \cdot b \ b \cdot a & b \cdot b \end{matrix} ight| = (a \cdot a)(b \cdot b) - (a \cdot b)(b \cdot a)$.Hence,the correct option is $B$.

If $a$ and $b$ are two vectors,then $(a \times b)^2$ equals

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