The angle between the planes x+y+2z=6 and 2x- | vedclass

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(C) The equations of the given planes are $x+y+2z-6=0$ and $2x-y+z-9=0$. Comparing these with the general form $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c

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$\frac{\pi}{3}$

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(C) The equations of the given planes are $x+y+2z-6=0$ and $2x-y+z-9=0$. Comparing these with the general form $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$, we get the normal vectors $\vec{n_1} = (1, 1, 2)$ and $\vec{n_2} = (2, -1, 1)$. The angle $	heta$ between the two planes is given by the formula: $\cos 	heta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$. Substituting the values: $\cos 	heta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+(-1)^2+1^2}}$. $\cos 	heta = \frac{|2 - 1 + 2|}{\sqrt{1+1+4} \sqrt{4+1+1}} = \frac{3}{\sqrt{6} 	imes \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Since $\cos 	heta = \frac{1}{2}$, we have $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

The angle between the planes $x+y+2z=6$ and $2x-y+z=9$ is:

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