Find the angle between the line \(\bar{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})\) and the plane \(\bar{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 8\).

Answer: \(\theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)\)

The line is parallel to \(\bar{b} = \hat{i} + \hat{j} + \hat{k}\) and the plane has normal \(\bar{n} = 2\hat{i} + \hat{j} + \hat{k}\).

If \(\theta\) is the angle between the line and the plane, \(\sin\theta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}||\bar{n}|}\).

\(\bar{b} \cdot \bar{n} = 2 + 1 + 1 = 4\), \(|\bar{b}| = \sqrt{3}\), \(|\bar{n}| = \sqrt{6}\).

\(\sin\theta = \frac{4}{\sqrt{3}\sqrt{6}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}\), so \(\theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)\).