Find the angle between the two planes 3x - 6y | vedclass

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(A) The equations of the planes are $3x - 6y + 2z - 7 = 0$ and $2x + 2y - 2z - 5 = 0$.Comparing these with $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x +

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$	heta = \cos^{-1}\left(\frac{5\sqrt{3}}{21}ight)$

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(A) The equations of the planes are $3x - 6y + 2z - 7 = 0$ and $2x + 2y - 2z - 5 = 0$.Comparing these with $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$,we get:$A_1 = 3, B_1 = -6, C_1 = 2$$A_2 = 2, B_2 = 2, C_2 = -2$The angle $	heta$ between the two planes is given by:$\cos 	heta = \left| \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} ight|$$\cos 	heta = \left| \frac{(3)(2) + (-6)(2) + (2)(-2)}{\sqrt{3^2 + (-6)^2 + 2^2} \sqrt{2^2 + 2^2 + (-2)^2}} ight|$$\cos 	heta = \left| \frac{6 - 12 - 4}{\sqrt{9 + 36 + 4} \sqrt{4 + 4 + 4}} ight|$$\cos 	heta = \left| \frac{-10}{\sqrt{49} \sqrt{12}} ight| = \left| \frac{-10}{7 	imes 2\sqrt{3}} ight| = \frac{10}{14\sqrt{3}} = \frac{5}{7\sqrt{3}}$Rationalizing the denominator,we get $\frac{5\sqrt{3}}{7 	imes 3} = \frac{5\sqrt{3}}{21}$.Therefore,$	heta = \cos^{-1}\left(\frac{5\sqrt{3}}{21}ight)$.

Find the angle between the two planes $3x - 6y + 2z = 7$ and $2x + 2y - 2z = 5$.

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