The angle theta between the line x - 1 = frac | vedclass

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(D) The line is given by $\frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{2}$. The direction vector of the line is $\vec{b} = \hat{i} - \hat{j} + 2\

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$\frac{\pi}{6}$

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(D) The line is given by $\frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{2}$. The direction vector of the line is $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.The normal vector to the plane is $\vec{n} = 2\hat{i} + \hat{j} + \hat{k}$.The angle $	heta$ between a line and a plane is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculate the dot product: $\vec{b} \cdot \vec{n} = (1)(2) + (-1)(1) + (2)(1) = 2 - 1 + 2 = 3$.Calculate the magnitudes: $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}$ and $|\vec{n}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$.Substitute the values: $\sin 	heta = \frac{|3|}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.Therefore, $	heta = \sin^{-1}\left(\frac{1}{2}ight) = 30^{\circ} = \frac{\pi}{6}$.

The angle $\theta$ between the line $x - 1 = \frac{y - 2}{-1} = \frac{z - 3}{2}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is

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