The combined equation for a pair of planes is | vedclass

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Question

(A) The given equation is $S = 2 x^2 - 6 y^2 - 12 z^2 + 18 y z + 2 z x + x y = 0$. Factorizing the quadratic form, we get $S = (x + 2 y - 2 z)(2 x - 3

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$\cos ^{-1}\left(\frac{16}{21}\right)$

Vedclass · Answer

(A) The given equation is $S = 2 x^2 - 6 y^2 - 12 z^2 + 18 y z + 2 z x + x y = 0$. Factorizing the quadratic form, we get $S = (x + 2 y - 2 z)(2 x - 3 y + 6 z) = 0$. Thus, the two planes are $P_1: x + 2 y - 2 z = 0$ and $P_2: 2 x - 3 y + 6 z = 0$. The normal vectors to these planes are $\vec{n_1} = (1, 2, -2)$ and $\vec{n_2} = (2, -3, 6)$. The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(-3) + (-2)(6) = 2 - 6 - 12 = -16$. Calculating the magnitudes: $|\vec{n_1}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$ and $|\vec{n_2}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7$. Therefore, $\cos 	heta = \frac{|-16|}{3 	imes 7} = \frac{16}{21}$. Hence, the acute angle is $	heta = \cos ^{-1}\left(\frac{16}{21}ight)$.

The combined equation for a pair of planes is $S \equiv 2 x^2-6 y^2-12 z^2+18 y z+2 z x+x y=0$. If one of the planes is parallel to $x+2 y-2 z=5$, then the acute angle between the planes $S=0$ is

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