The angle theta between the line vec{r} = (ha | vedclass

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(A) The line is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$.The plane is given by $\vec{r} \cdot \ve

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$\sin^{-1} \left( \frac{\sqrt{2}}{3} \right)$

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(A) The line is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$.The plane is given by $\vec{r} \cdot \vec{n} = d$, where $\vec{n} = 2\hat{i} - \hat{j} + \hat{k}$.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: $\vec{b} \cdot \vec{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.Calculate the magnitudes: $|\vec{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.Substitute into the formula: $\sin 	heta = \frac{|2|}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$.Therefore, $	heta = \sin^{-1} \left( \frac{\sqrt{2}}{3} ight)$.

The angle $\theta$ between the line $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 8$ is

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