For vectors vec{a} and vec{b},if |vec{a}|=3,| | vedclass

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(C) Given that $|\vec{a}| = 3$,$|\vec{b}| = \frac{\sqrt{2}}{3}$,and $|\vec{a} 	imes \vec{b}| = 1$ (since it is a unit vector).We know the formula for

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$\frac{\pi}{4}$

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(C) Given that $|\vec{a}| = 3$,$|\vec{b}| = \frac{\sqrt{2}}{3}$,and $|\vec{a} 	imes \vec{b}| = 1$ (since it is a unit vector).We know the formula for the magnitude of the cross product: $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin(	heta)$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.Substituting the given values: $1 = 3 	imes \frac{\sqrt{2}}{3} 	imes \sin(	heta)$.This simplifies to: $1 = \sqrt{2} \sin(	heta)$.Therefore,$\sin(	heta) = \frac{1}{\sqrt{2}}$.Since $\sin(	heta) = \frac{1}{\sqrt{2}}$,the angle $	heta = \frac{\pi}{4}$.

For vectors $\vec{a}$ and $\vec{b}$,if $|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

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