The angle between the pair of lines given by $\vec r= \hat i+2\hat j3-3\hat k+λ (\hat i−2\hat j+2\hat k)$ and $\vec r= 5\hat i+\hat j+\hat k+μ (3\hat i-2\hat j +6\hat k)$

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

The correct answer is Option (3) → $\cos^{-1}(\frac{19}{21})$

Direction vector of first line: $\vec{d_1} = \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$

Direction vector of second line: $\vec{d_2} = \mu(3\hat{i} - 2\hat{j} + 6\hat{k})$

Let $\vec{a} = (1, -2, 2)$ and $\vec{b} = (3, -2, 6)$

Angle $\theta$ between the lines is given by:

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

$\vec{a} \cdot \vec{b} = 1\cdot3 + (-2)\cdot(-2) + 2\cdot6 = 3 + 4 + 12 = 19$

$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

$|\vec{b}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

$\cos\theta = \frac{19}{3 \cdot 7} = \frac{19}{21}$

Angle between the lines is $\theta = \cos^{-1}\left(\frac{19}{21}\right)$