Let u and v be two vectors. Then |u-v|=||u|-| | vedclass

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(B) We know that $|u-v|^2 = |u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v})$.Also,$(||u|-|v||)^2 = |u|^2 + |v|^2 - 2|u||v|$.For the equality $|u-v| = ||u|-|v

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$u$ and $v$ have the same direction

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(B) We know that $|u-v|^2 = |u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v})$.Also,$(||u|-|v||)^2 = |u|^2 + |v|^2 - 2|u||v|$.For the equality $|u-v| = ||u|-|v||$ to hold,their squares must be equal:$|u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v}) = |u|^2 + |v|^2 - 2|u||v|$.This simplifies to $\vec{u} \cdot \vec{v} = |u||v|$.Since $\vec{u} \cdot \vec{v} = |u||v| \cos 	heta$,we have $|u||v| \cos 	heta = |u||v|$.This implies $\cos 	heta = 1$,which means $	heta = 0$.Therefore,$u$ and $v$ must have the same direction.

Let $u$ and $v$ be two vectors. Then $|u-v|=||u|-|v||$ if and only if

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