A plane pi_1 contains the vectors bar{i}+bar{ | vedclass

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(C) The normal vector $\bar{n}_1$ to plane $\pi_1$ is given by $(\bar{i}+\bar{j}) 	imes (\bar{i}+2\bar{j}) = \bar{k}$.The normal vector $\bar{n}_2$ t

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$\cos^{-1}\left(\frac{4}{3\sqrt{5}}\right)$

Vedclass · Answer

(C) The normal vector $\bar{n}_1$ to plane $\pi_1$ is given by $(\bar{i}+\bar{j}) 	imes (\bar{i}+2\bar{j}) = \bar{k}$.The normal vector $\bar{n}_2$ to plane $\pi_2$ is given by $(2\bar{i}-\bar{j}) 	imes (3\bar{i}+2\bar{k}) = -2\bar{i}-4\bar{j}+3\bar{k}$.The vector $\bar{a}$ parallel to the line of intersection is $\bar{n}_1 	imes \bar{n}_2 = \bar{k} 	imes (-2\bar{i}-4\bar{j}+3\bar{k}) = 4\bar{i}-2\bar{j}$.Let $\bar{b} = \bar{i}-2\bar{j}+2\bar{k}$.The angle $	heta$ between $\bar{a}$ and $\bar{b}$ is given by $\cos 	heta = \frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}| |\bar{b}|}$.$\bar{a} \cdot \bar{b} = (4)(1) + (-2)(-2) + (0)(2) = 4+4 = 8$.$|\bar{a}| = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.$|\bar{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = 3$.$\cos 	heta = \frac{8}{(2\sqrt{5})(3)} = \frac{8}{6\sqrt{5}} = \frac{4}{3\sqrt{5}}$.Thus, $	heta = \cos^{-1}\left(\frac{4}{3\sqrt{5}}ight)$.

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