Let vec{a} and vec{b} be two non-collinear ve | vedclass

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(A) Given,$|\vec{a}|=|\vec{b}|=1$. Since $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$,let $	heta$ be the angle between $\vec{a}$ and $\vec{b}$.

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

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(A) Given,$|\vec{a}|=|\vec{b}|=1$. Since $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$,let $	heta$ be the angle between $\vec{a}$ and $\vec{b}$. Then $\vec{a} \cdot \vec{b} = \cos 	heta$. So,$\vec{u} = \vec{a} - \cos 	heta \vec{b}$. Calculating the magnitude squared: $|\vec{u}|^2 = |\vec{a}|^2 + \cos^2 	heta |\vec{b}|^2 - 2 \cos 	heta (\vec{a} \cdot \vec{b}) = 1 + \cos^2 	heta - 2 \cos^2 	heta = 1 - \cos^2 	heta = \sin^2 	heta$. Thus,$|\vec{u}| = \sin 	heta$ (since $\sin 	heta > 0$ for non-collinear vectors). Also,$|\vec{v}| = |\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta = \sin 	heta$. Therefore,$|\vec{v}| = |\vec{u}|$. Now,$\vec{u} \cdot \vec{v} = (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot (\vec{a} 	imes \vec{b}) = \vec{a} \cdot (\vec{a} 	imes \vec{b}) - (\vec{a} \cdot \vec{b}) \vec{b} \cdot (\vec{a} 	imes \vec{b}) = 0 - 0 = 0$. Hence,$|\vec{v}| = |\vec{u}| + |\vec{u} \cdot \vec{v}| = |\vec{u}| + 0 = |\vec{u}|$.

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors of unit modulus. If $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v}=\vec{a} \times \vec{b}$,then $|\vec{v}|=$

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