Let vec{u}=hat{i}-hat{j}-2hat{k},vec{v}=2hat{ | vedclass

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(A) Given $\vec{u}=(1, -1, -2)$,$\vec{v}=(2, 1, -1)$,and $\vec{v} \cdot \vec{w}=2$. We are given the equation $\vec{v} 	imes \vec{w} = \vec{u} + \lam

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(A) Given $\vec{u}=(1, -1, -2)$,$\vec{v}=(2, 1, -1)$,and $\vec{v} \cdot \vec{w}=2$. We are given the equation $\vec{v} 	imes \vec{w} = \vec{u} + \lambda\vec{v} \quad \dots(1)$. Taking the dot product of equation $(1)$ with $\vec{v}$: $\vec{v} \cdot (\vec{v} 	imes \vec{w}) = \vec{v} \cdot \vec{u} + \lambda(\vec{v} \cdot \vec{v})$. Since $\vec{v} \cdot (\vec{v} 	imes \vec{w}) = 0$,we have $0 = (2 - 1 + 2) + \lambda(2^2 + 1^2 + (-1)^2)$. $0 = 3 + 6\lambda \implies \lambda = -\frac{1}{2}$. Now,taking the dot product of equation $(1)$ with $\vec{w}$: $\vec{w} \cdot (\vec{v} 	imes \vec{w}) = \vec{w} \cdot \vec{u} + \lambda(\vec{w} \cdot \vec{v})$. Since $\vec{w} \cdot (\vec{v} 	imes \vec{w}) = 0$,we have $0 = \vec{u} \cdot \vec{w} + \lambda(2)$. $\vec{u} \cdot \vec{w} = -2\lambda = -2(-\frac{1}{2}) = 1$.

Let $\vec{u}=\hat{i}-\hat{j}-2\hat{k}$,$\vec{v}=2\hat{i}+\hat{j}-\hat{k}$,$\vec{v} \cdot \vec{w}=2$ and $\vec{v} \times \vec{w}=\vec{u}+\lambda\vec{v}$. Then $\vec{u} \cdot \vec{w}$ is equal to $......$

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