For vectors bar{a} and bar{b},|bar{a}| = frac | vedclass

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Question

(A) We know that the magnitude of the cross product of two vectors $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} 	imes \bar{b}| = |\bar{a}| |\bar{b}|

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) We know that the magnitude of the cross product of two vectors $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} 	imes \bar{b}| = |\bar{a}| |\bar{b}| \sin(	heta)$,where $	heta$ is the angle between the vectors.Given that $|\bar{a}| = \frac{2}{3}$,$|\bar{b}| = 3$,and $|\bar{a} 	imes \bar{b}| = 1$.Substituting these values into the formula:$1 = (\frac{2}{3}) 	imes 3 	imes \sin(	heta)$$1 = 2 \sin(	heta)$$\sin(	heta) = \frac{1}{2}$Since $\sin(	heta) = \frac{1}{2}$,the angle $	heta$ is $\frac{\pi}{6}$ (or $30^{\circ}$).Therefore,the correct option is $A$.

For vectors $\bar{a}$ and $\bar{b}$,$|\bar{a}| = \frac{2}{3}$,$|\bar{b}| = 3$ and $|\bar{a} \times \bar{b}| = 1$,then the angle between $\bar{a}$ and $\bar{b}$ is . . . . . . .

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