Let pi_1 be a plane passing through the point | vedclass

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(A) The normal vector to plane $\pi_1$ is $\vec{n}_1 = 0\hat{i} - 1\hat{j} + 2\hat{k}$.The line $L$ passes through points $A(3, -2, 1)$ and $B(-1, 3,

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$\sqrt{\frac{5}{41}}$

Vedclass · Answer

(A) The normal vector to plane $\pi_1$ is $\vec{n}_1 = 0\hat{i} - 1\hat{j} + 2\hat{k}$.The line $L$ passes through points $A(3, -2, 1)$ and $B(-1, 3, 1)$. The direction vector of line $L$ is $\vec{v} = (-1-3)\hat{i} + (3-(-2))\hat{j} + (1-1)\hat{k} = -4\hat{i} + 5\hat{j} + 0\hat{k}$.Since line $L$ is normal to plane $\pi_2$, the normal vector to plane $\pi_2$ is $\vec{n}_2 = -4\hat{i} + 5\hat{j} + 0\hat{k}$.The angle $	heta$ between the two planes is given by $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$.$\vec{n}_1 \cdot \vec{n}_2 = (0)(-4) + (-1)(5) + (2)(0) = -5$.$|\vec{n}_1| = \sqrt{0^2 + (-1)^2 + 2^2} = \sqrt{5}$.$|\vec{n}_2| = \sqrt{(-4)^2 + 5^2 + 0^2} = \sqrt{16 + 25} = \sqrt{41}$.$\cos 	heta = \frac{|-5|}{\sqrt{5} \cdot \sqrt{41}} = \frac{5}{\sqrt{205}} = \frac{5}{\sqrt{5} \cdot \sqrt{41}} = \sqrt{\frac{5}{41}}$.

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