If alpha is the acute angle between the plane | vedclass

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(A) The general equation of a pair of planes passing through the origin is $ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0$.Comparing this with the given

Vedclass · Accepted Answer

$\frac{16}{21}$

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(A) The general equation of a pair of planes passing through the origin is $ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0$.Comparing this with the given equation $2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0$,we have $a=2, b=-6, c=-12, f=9, g=1, h=0.5$.The planes are represented by $2x^2 + xy - 6y^2 + 2zx + 18yz - 12z^2 = 0$.Factoring the quadratic expression,we get $(x + 2y - 2z)(2x - 3y + 6z) = 0$.Thus,the equations of the planes are $P_1: x + 2y - 2z = 0$ and $P_2: 2x - 3y + 6z = 0$.The normal vectors to these planes are $\vec{n_1} = (1, 2, -2)$ and $\vec{n_2} = (2, -3, 6)$.The cosine of the angle $\alpha$ between the planes is given by $\cos \alpha = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} ight|$.$\vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(-3) + (-2)(6) = 2 - 6 - 12 = -16$.$|\vec{n_1}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3$.$|\vec{n_2}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = 7$.Therefore,$\cos \alpha = \left| \frac{-16}{3 	imes 7} ight| = \frac{16}{21}$.

If $\alpha$ is the acute angle between the planes $P_1$ and $P_2$,where the combined equation of the planes $P_1$ and $P_2$ is $2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0$,then the value of $\cos \alpha$ is:

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