The angle between the line vec{r} = (hat{i} + | vedclass

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Question

(A) The direction vector of the line is $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$.The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} + 3\ha

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$\cos^{-1} (\frac{4}{\sqrt{66}})$

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(A) The direction vector of the line is $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$.The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} + 3\hat{k}$.The angle $	heta$ between the line and the normal to the plane is given by the formula $\cos 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(1) + (-1)(1) + (1)(3) = 2 - 1 + 3 = 4$.Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n}| = \sqrt{1^2 + 1^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}$.Thus,$\cos 	heta = \frac{4}{\sqrt{6} \cdot \sqrt{11}} = \frac{4}{\sqrt{66}}$.Therefore,$	heta = \cos^{-1} (\frac{4}{\sqrt{66}})$.

The angle between the line $\vec{r} = (\hat{i} + \hat{j} - 2\hat{k}) + \lambda (2\hat{i} - \hat{j} + \hat{k})$ and the normal to the plane $\vec{r} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 2$ is:

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