Find the angle between the pair of lines given by $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda (\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{j} + \mu (3\hat{i} + 2\hat{j} + 6\hat{k})$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\cos^{-1} \left( \frac{19}{21} \right)$ ##
Here $\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
The angle $\theta$ between the two lines is given by
$\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| \, |\vec{b_2}|}$
$= \frac{|(\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (3\hat{i} + 2\hat{j} + 6\hat{k})|}{\sqrt{1+4+4} \sqrt{9+4+36}}$
$= \frac{|3 + 4 + 12|}{\sqrt{9} \sqrt{49}} = \frac{19}{3 \times 7} = \frac{19}{21}$
Hence $\theta = \cos^{-1} \left( \frac{19}{21} \right)$