Let pi_1 be the plane determined by the vecto | vedclass

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(B) The normal vector $\vec{n}_1$ to plane $\pi_1$ is given by the cross product of its spanning vectors: $\vec{n}_1 = (\hat{i}+\hat{j}) 	imes (\hat{

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$2$

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(B) The normal vector $\vec{n}_1$ to plane $\pi_1$ is given by the cross product of its spanning vectors: $\vec{n}_1 = (\hat{i}+\hat{j}) 	imes (\hat{j}+\hat{k}) = \hat{i}-\hat{j}+\hat{k}$.The normal vector $\vec{n}_2$ to plane $\pi_2$ is given by the cross product of its spanning vectors: $\vec{n}_2 = (\hat{i}-\hat{j}) 	imes (\hat{i}+\hat{j}-\hat{k}) = \hat{i}+\hat{j}+2\hat{k}$.The vector $\vec{b}$ parallel to the line of intersection is $\vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & 1 & 2 \end{vmatrix} = -3\hat{i}-\hat{j}+2\hat{k}$.Since $\vec{a}$ is parallel to $\vec{b}$,we have $\vec{a} = \lambda(3\hat{i}+\hat{j}-2\hat{k})$.Given $|\vec{a}| = \sqrt{14}$,we have $|\lambda| \sqrt{3^2+1^2+(-2)^2} = \sqrt{14} \Rightarrow |\lambda| \sqrt{14} = \sqrt{14} \Rightarrow \lambda = \pm 1$.Thus,$\vec{a} = \pm(3\hat{i}+\hat{j}-2\hat{k})$.Finally,$|\vec{a} \cdot (\hat{i}+\hat{j}+\hat{k})| = |\pm(3+1-2)| = |\pm 2| = 2$.

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