The angle between two planes x + 2y + 2z = 3 | vedclass

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(A) The angle $	heta$ between two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ is given by the formula:$\cos 	heta = \frac{|a_1a

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$\cos^{-1} \frac{3\sqrt{2}}{10}$

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(A) The angle $	heta$ between two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ is given by the formula:$\cos 	heta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$Here,the normal vectors are $\vec{n_1} = (1, 2, 2)$ and $\vec{n_2} = (-5, 3, 4)$.Substituting the values:$\cos 	heta = \frac{|(1)(-5) + (2)(3) + (2)(4)|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{(-5)^2 + 3^2 + 4^2}}$$\cos 	heta = \frac{|-5 + 6 + 8|}{\sqrt{1 + 4 + 4} \sqrt{25 + 9 + 16}}$$\cos 	heta = \frac{9}{\sqrt{9} \sqrt{50}}$$\cos 	heta = \frac{9}{3 	imes 5\sqrt{2}} = \frac{3}{5\sqrt{2}}$Rationalizing the denominator:$\cos 	heta = \frac{3 	imes \sqrt{2}}{5 	imes 2} = \frac{3\sqrt{2}}{10}$Therefore,$	heta = \cos^{-1} \left( \frac{3\sqrt{2}}{10} ight)$.

The angle between two planes $x + 2y + 2z = 3$ and $-5x + 3y + 4z = 9$ is

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