Find the angle between the planes x + y + 2z | vedclass

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(C) The angle $	heta$ between two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by the formula:$\cos 	heta = \fr

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$\pi/3$

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(C) The angle $	heta$ between two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by the formula:$\cos 	heta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$For the given planes $x + y + 2z = 9$ and $2x - y + z = 15$,we have:$a_1 = 1, b_1 = 1, c_1 = 2$$a_2 = 2, b_2 = -1, c_2 = 1$Substituting these values into the formula:$\cos 	heta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{2^2 + (-1)^2 + 1^2}}$$\cos 	heta = \frac{|2 - 1 + 2|}{\sqrt{1 + 1 + 4} \sqrt{4 + 1 + 1}}$$\cos 	heta = \frac{3}{\sqrt{6} \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = \frac{\pi}{3}$.

Find the angle between the planes $x + y + 2z = 9$ and $2x - y + z = 15$.

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