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

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

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

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(A) The angle $	heta$ between a line with direction vector $\bar{b}$ and a plane with normal vector $\bar{n}$ is given by $\sin 	heta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.Given the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$,the direction vector is $\bar{b} = \hat{i}+\hat{j}+\hat{k}$.Given the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$,the normal vector is $\bar{n} = 2\hat{i}-\hat{j}+\hat{k}$.Calculate the magnitudes: $|\bar{b}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{4+1+1} = \sqrt{6}$.Calculate the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.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 acute angle between the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$ is

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