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(D) The normal vectors to the two planes are $\vec{n}_1 = (\hat{i}+\hat{j}) 	imes (\hat{i}+\hat{k}) = \hat{i}-\hat{j}-\hat{k}$ and $\vec{n}_2 = (\hat

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$(\frac{\pi}{4}, 6)$

Vedclass · Answer

(D) The normal vectors to the two planes are $\vec{n}_1 = (\hat{i}+\hat{j}) 	imes (\hat{i}+\hat{k}) = \hat{i}-\hat{j}-\hat{k}$ and $\vec{n}_2 = (\hat{i}-\hat{j}) 	imes (\hat{j}-\hat{k}) = \hat{i}+\hat{j}+\hat{k}$.Since $\vec{a}$ is parallel to the line of intersection,$\vec{a} = \lambda(\vec{n}_1 	imes \vec{n}_2)$.$\vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -1 \ 1 & 1 & 1 \end{vmatrix} = 0\hat{i} - 2\hat{j} + 2\hat{k} = -2\hat{j} + 2\hat{k}$.So,$\vec{a} = \lambda(-2\hat{j} + 2\hat{k})$.Given $\vec{a} \cdot \vec{b} = 6$,where $\vec{b} = 2\hat{i}-2\hat{j}+\hat{k}$.$\lambda(-2\hat{j} + 2\hat{k}) \cdot (2\hat{i}-2\hat{j}+\hat{k}) = \lambda(0 + 4 + 2) = 6\lambda = 6 \implies \lambda = 1$.Thus,$\vec{a} = -2\hat{j} + 2\hat{k}$.$|\vec{a}| = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ and $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$.$\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{6}{2\sqrt{2} 	imes 3} = \frac{1}{\sqrt{2}} \implies 	heta = \frac{\pi}{4}$.Using the identity $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = (8)(9) - 6^2 = 72 - 36 = 36$.Therefore,$|\vec{a} 	imes \vec{b}| = 6$.The ordered pair is $(\frac{\pi}{4}, 6)$.

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