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(A) Given that $2\vec{u} 	imes 3\vec{v}$ is a unit vector,its magnitude must be $1$.Thus,$|2\vec{u} 	imes 3\vec{v}| = 1$.Since $\vec{u}$ and $\vec{v

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exactly one value of $	heta$

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(A) Given that $2\vec{u} 	imes 3\vec{v}$ is a unit vector,its magnitude must be $1$.Thus,$|2\vec{u} 	imes 3\vec{v}| = 1$.Since $\vec{u}$ and $\vec{v}$ are unit vectors,we have $|\vec{u}| = 1$ and $|\vec{v}| = 1$.Using the property of the cross product,$|2\vec{u} 	imes 3\vec{v}| = 6|\vec{u}||\vec{v}|\sin	heta = 6(1)(1)\sin	heta = 6\sin	heta$.According to the problem,$6\sin	heta = 1$,which implies $\sin	heta = \frac{1}{6}$.Since $	heta$ is an acute angle,there exists exactly one value of $	heta$ such that $	heta = \arcsin(\frac{1}{6})$.

If $\vec{u}$ and $\vec{v}$ are unit vectors and $\theta$ is the acute angle between them,then $2\vec{u} \times 3\vec{v}$ is a unit vector for

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