vec{a}=hat{i}+hat{j}-2 hat{k}, vec{b}=hat{i}- | vedclass

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(C) Since $\vec{d}$ is normal to the plane of $\vec{a}$ and $\vec{b}$, it must be parallel to $\vec{a} 	imes \vec{b}$.Thus, $\vec{d} = \lambda(\vec{a

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$\sqrt{3}$

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(C) Since $\vec{d}$ is normal to the plane of $\vec{a}$ and $\vec{b}$, it must be parallel to $\vec{a} 	imes \vec{b}$.Thus, $\vec{d} = \lambda(\vec{a} 	imes \vec{b})$.First, calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -2 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 - 4) - \hat{j}(1 - (-2)) + \hat{k}(-2 - 1) = -3\hat{i} - 3\hat{j} - 3\hat{k}$.We can write $\vec{d} = \mu(\hat{i} + \hat{j} + \hat{k})$ where $\mu = -3\lambda$.Given $\vec{d} \cdot \vec{c} = 2$, we substitute $\vec{c} = 2\hat{i} + \hat{j} - \hat{k}$:$\mu(\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{i} + \hat{j} - \hat{k}) = 2$$\mu(2 + 1 - 1) = 2 \Rightarrow 2\mu = 2 \Rightarrow \mu = 1$.Therefore, $\vec{d} = \hat{i} + \hat{j} + \hat{k}$.The magnitude is $|\vec{d}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

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