Find the angle between the following two lines: $\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$; $\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$.

$\cos^{-1}\left(\frac{19}{21}\right)$

The correct answer is Option (2) → $\cos^{-1}\left(\frac{19}{21}\right)$ ##

$\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$

$\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$

$\vec{b} = 3\hat{i} + 2\hat{j} + 6\hat{k}$

$\vec{c} = \hat{i} + 2\hat{j} + 2\hat{k}$

$∴\cos\theta = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{b}| |\vec{c}|}$

$= \frac{|(3\hat{i} + 2\hat{j} + 6\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k})|}{\sqrt{3^2 + 2^2 + 6^2} \sqrt{1^2 + 2^2 + 2^2}}$

$\cos\theta = \frac{3 + 4 + 12}{7 \times 3} \Rightarrow \cos\theta = \frac{19}{21}$

$\theta = \cos^{-1}\left(\frac{19}{21}\right)$