vec{u}, vec{v}, vec{w} are three unit vectors | vedclass

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(B) Given that $|\vec{u}|=|\vec{v}|=|\vec{w}|=1$. $\vec{p} \cdot \vec{u}=(\vec{u}+\vec{v}+\vec{w}) \cdot \vec{u}=\frac{3}{2} \Rightarrow |\vec{u}|^2+\

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$2$

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(B) Given that $|\vec{u}|=|\vec{v}|=|\vec{w}|=1$. $\vec{p} \cdot \vec{u}=(\vec{u}+\vec{v}+\vec{w}) \cdot \vec{u}=\frac{3}{2} \Rightarrow |\vec{u}|^2+\vec{u} \cdot \vec{v}+\vec{w} \cdot \vec{u}=\frac{3}{2} \Rightarrow \vec{u} \cdot \vec{v}+\vec{w} \cdot \vec{u}=\frac{1}{2}$ ....$(i)$ $\vec{p} \cdot \vec{v}=(\vec{u}+\vec{v}+\vec{w}) \cdot \vec{v}=\frac{7}{4} \Rightarrow \vec{u} \cdot \vec{v}+|\vec{v}|^2+\vec{v} \cdot \vec{w}=\frac{7}{4} \Rightarrow \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}=\frac{3}{4}$ ....$(ii)$ $|\vec{p}|^2=|\vec{u}+\vec{v}+\vec{w}|^2=4 \Rightarrow |\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=4 \Rightarrow \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}=\frac{1}{2}$ ....$(iii)$ Solving $(i)$,$(ii)$,and $(iii)$: Subtracting $(ii)$ from $(iii)$: $\vec{w} \cdot \vec{u} = \frac{1}{2} - \frac{3}{4} = -\frac{1}{4}$. Subtracting $(i)$ from $(iii)$: $\vec{v} \cdot \vec{w} = \frac{1}{2} - \frac{1}{2} = 0$. Then $\vec{u} \cdot \vec{v} = \frac{1}{2} - (-\frac{1}{4}) = \frac{3}{4}$. Given $\vec{v}=K \vec{q}=K[\vec{u} 	imes(\vec{v} 	imes \vec{w})]=K[(\vec{u} \cdot \vec{w}) \vec{v}-(\vec{u} \cdot \vec{v}) \vec{w}]$. Substituting values: $\vec{v}=K[-\frac{1}{4} \vec{v}-\frac{3}{4} \vec{w}] \Rightarrow \vec{v} = -\frac{K}{4} \vec{v} - \frac{3K}{4} \vec{w} \Rightarrow (1+\frac{K}{4}) \vec{v} = -\frac{3K}{4} \vec{w}$. Taking magnitudes: $|1+\frac{K}{4}| = |-\frac{3K}{4}| \Rightarrow |4+K| = |3K|$. $4+K = 3K \Rightarrow 2K=4 \Rightarrow K=2$ or $4+K = -3K \Rightarrow 4K=-4 \Rightarrow K=-1$. Checking the vector equation: $(1+\frac{K}{4}) \vec{v} = -\frac{3K}{4} \vec{w}$. If $K=-1$,$\frac{3}{4} \vec{v} = \frac{3}{4} \vec{w} \Rightarrow \vec{v}=\vec{w}$,but $\vec{v} \cdot \vec{w}=0$,which is impossible. Thus,$K=2$.

$\vec{u}, \vec{v}, \vec{w}$ are three unit vectors. Let $\vec{p}=\vec{u}+\vec{v}+\vec{w}$ and $\vec{q}=\vec{u} \times(\vec{v} \times \vec{w})$. If $\vec{p} \cdot \vec{u}=\frac{3}{2}, \vec{p} \cdot \vec{v}=\frac{7}{4}, |\vec{p}|=2$ and $\vec{v}=K \vec{q}$,then $K=$

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