Let bar{u}=hat{i}+hat{j},bar{v}=hat{i}-hat{j} | vedclass

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(D) Given $\bar{u} \cdot \hat{n}=0$ and $\bar{v} \cdot \hat{n}=0$. This implies that the unit vector $\hat{n}$ is perpendicular to both $\bar{u}$ and

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$3$

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(D) Given $\bar{u} \cdot \hat{n}=0$ and $\bar{v} \cdot \hat{n}=0$. This implies that the unit vector $\hat{n}$ is perpendicular to both $\bar{u}$ and $\bar{v}$. Therefore,$\hat{n}$ is parallel to the cross product $\bar{u} 	imes \bar{v}$. Thus,$\hat{n} = \pm \frac{\bar{u} 	imes \bar{v}}{|\bar{u} 	imes \bar{v}|}$. First,calculate the cross product: $\bar{u} 	imes \bar{v} = (\hat{i}+\hat{j}) 	imes (\hat{i}-\hat{j}) = \hat{i} 	imes \hat{i} - \hat{i} 	imes \hat{j} + \hat{j} 	imes \hat{i} - \hat{j} 	imes \hat{j} = 0 - \hat{k} - \hat{k} - 0 = -2\hat{k}$. The magnitude is $|\bar{u} 	imes \bar{v}| = |-2\hat{k}| = 2$. So,$\hat{n} = \pm \frac{-2\hat{k}}{2} = \pm \hat{k}$. Now,calculate $|\bar{w} \cdot \hat{n}|$: $|\bar{w} \cdot \hat{n}| = |(\hat{i}+2\hat{j}+3\hat{k}) \cdot (\pm \hat{k})| = |\pm 3| = 3$.

Let $\bar{u}=\hat{i}+\hat{j}$,$\bar{v}=\hat{i}-\hat{j}$ and $\bar{w}=\hat{i}+2\hat{j}+3\hat{k}$. If $\hat{n}$ is a unit vector such that $\bar{u} \cdot \hat{n}=0$ and $\bar{v} \cdot \hat{n}=0$,then $|\bar{w} \cdot \hat{n}|$ is equal to

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