Find the angle between the planes x+2y+2z-5=0 | vedclass

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(B) The equations of the given planes are $x+2y+2z-5=0$ and $3x+3y+2z-8=0$. The normal vectors to these planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2

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$\cos^{-1}\left(\frac{13}{3\sqrt{22}}\right)$

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(B) The equations of the given planes are $x+2y+2z-5=0$ and $3x+3y+2z-8=0$. The normal vectors to these planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + 3\hat{j} + 2\hat{k}$. The angle $	heta$ between two planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (1)(3) + (2)(3) + (2)(2) = 3 + 6 + 4 = 13$. Calculating the magnitudes: $|\vec{n_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$ and $|\vec{n_2}| = \sqrt{3^2 + 3^2 + 2^2} = \sqrt{9+9+4} = \sqrt{22}$. Thus,$\cos 	heta = \frac{13}{3\sqrt{22}}$. Therefore,$	heta = \cos^{-1}\left(\frac{13}{3\sqrt{22}}ight)$.

Find the angle between the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$.

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