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

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(A) The given planes are $2x - y + z = 6$ and $x + y + 2z = 3$. The normal vectors to these planes are $\vec{n}_1 = 2\hat{i} - \hat{j} + \hat{k}$ and

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$\frac{\pi}{3}$

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(A) The given planes are $2x - y + z = 6$ and $x + y + 2z = 3$. The normal vectors to these planes are $\vec{n}_1 = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n}_2 = \hat{i} + \hat{j} + 2\hat{k}$. The angle $	heta$ between the two planes is given by the formula $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{||\vec{n}_1|| ||\vec{n}_2||}$. Calculating the dot product: $\vec{n}_1 \cdot \vec{n}_2 = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$. Calculating the magnitudes: $||\vec{n}_1|| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $||\vec{n}_2|| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$. Substituting these values into the formula: $\cos 	heta = \frac{|3|}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Since $\cos 	heta = \frac{1}{2}$,we have $	heta = \frac{\pi}{3}$.

The angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is

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