The acute angle between the planes 2x - y + z | vedclass

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(C) The normal vectors to the given planes $2x - y + z = 6$ and $x + y + 2z = 3$ are $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{

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$60$

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(C) The normal vectors to the given planes $2x - y + z = 6$ and $x + y + 2z = 3$ are $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$ respectively.The angle $	heta$ between the two planes is given by the formula $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.Calculate the dot product: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.Calculate the magnitudes: $|\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.Substitute the 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 = 60^o$.

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

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