Let theta be the angle between the planes P_1 | vedclass

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Question

(A) The normal vectors to the planes $P_1$ and $P_2$ are $\vec{n}_1 = \hat{i}+\hat{j}+2\hat{k}$ and $\vec{n}_2 = 2\hat{i}-\hat{j}+\hat{k}$ respectivel

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The normal vectors to the planes $P_1$ and $P_2$ are $\vec{n}_1 = \hat{i}+\hat{j}+2\hat{k}$ and $\vec{n}_2 = 2\hat{i}-\hat{j}+\hat{k}$ respectively.The angle $	heta$ between the two planes is given by $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{||\vec{n}_1|| ||\vec{n}_2||}$.$\cos 	heta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+(-1)^2+1^2}} = \frac{|2-1+2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.Thus,$	heta = \frac{\pi}{3}$.Let $L$ be a line making an angle $	heta$ with the normal of $P_2$. The angle $\alpha$ between the line $L$ and the plane $P_2$ is related to the angle $	heta$ between the line and the normal by $\alpha = \frac{\pi}{2} - 	heta$.Since $	heta = \frac{\pi}{3}$,we have $\alpha = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.We need to calculate $(	an^2 	heta)(\cot^2 \alpha)$.$(	an^2 \frac{\pi}{3})(\cot^2 \frac{\pi}{6}) = ((\sqrt{3})^2)((\sqrt{3})^2) = (3)(3) = 9$.

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