|a times b|^2 + (a cdot 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

Vedclass · Accepted Answer

$|a|^2 |b|^2$

Vedclass · Answer

(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.Squaring both sides,we get $|a 	imes b|^2 = |a|^2 |b|^2 \sin^2 	heta$.We also know that the dot product is given by $a \cdot b = |a| |b| \cos 	heta$.Squaring both sides,we get $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 	heta$.Adding these two expressions:$|a 	imes b|^2 + (a \cdot b)^2 = |a|^2 |b|^2 \sin^2 	heta + |a|^2 |b|^2 \cos^2 	heta$$= |a|^2 |b|^2 (\sin^2 	heta + \cos^2 	heta)$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:$|a 	imes b|^2 + (a \cdot b)^2 = |a|^2 |b|^2$.Since $|a|^2 = a \cdot a$ and $|b|^2 = b \cdot b$,the result is $(a \cdot a)(b \cdot b)$.

$|a \times b|^2 + (a \cdot b)^2 = ?$

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