Let vec{u} and vec{v} be two non-zero vectors | vedclass

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(C) The magnitude of the cross product of two vectors $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}| \sin 	heta$,

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$\vec{u} \cdot \vec{v}$

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(C) The magnitude of the cross product of two vectors $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}| \sin 	heta$,where $	heta$ is the angle between them.Given $	heta = 45^{\circ}$,we have $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}| \sin 45^{\circ} = |\vec{u}||\vec{v}| \left( \frac{1}{\sqrt{2}} ight)$.The dot product of the two vectors is given by $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos 	heta$.For $	heta = 45^{\circ}$,$\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos 45^{\circ} = |\vec{u}||\vec{v}| \left( \frac{1}{\sqrt{2}} ight)$.Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,it follows that $|\vec{u} 	imes \vec{v}| = \vec{u} \cdot \vec{v}$.

Let $\vec{u}$ and $\vec{v}$ be two non-zero vectors with the intermediate angle $45^{\circ}$. Then $|\vec{u} \times \vec{v}|=$

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