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

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(B) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left| \frac

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

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(B) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} ight|$.Here,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} + \hat{k}$.Calculating the dot product: $\vec{b} \cdot \vec{n} = (1)(2) + (1)(1) + (1)(1) = 2 + 1 + 1 = 4$.Calculating the magnitudes: $|\vec{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{n}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$.Substituting these values into the formula: $\sin 	heta = \left| \frac{4}{\sqrt{3} \cdot \sqrt{6}} ight| = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{4\sqrt{2}}{3 \cdot 2} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}ight)$.

The acute angle between the line $\vec{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\vec{r} \cdot(2\hat{i}+\hat{j}+\hat{k})=5$ is

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