The angle between two lines whose direction ratios are propotional to $1,1,-2$ and $(\sqrt{3}-1),(-\sqrt{3}-1),-4$ is :

We need to find the angle between two lines with direction ratios proportional to $(1, 1, -2)$ and $(\sqrt{3}-1, -\sqrt{3}-1, -4)$.

For any line with direction ratios $(a, b, c)$, the unit vector is obtained by dividing by the magnitude $\sqrt{a^2+b^2+c^2}$.

For the first line $(1, 1, -2)$:

Magnitude = $\sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

Unit vector $\vec{a} = \dfrac{1}{\sqrt{6}}(1, 1, -2)$

For the second line $(\sqrt{3}-1, -\sqrt{3}-1, -4)$:

Magnitude = $\sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + (-4)^2}$

Simplifying: $(\sqrt{3}-1)^2 = 4 - 2\sqrt{3}$, $(-\sqrt{3}-1)^2 = 4 + 2\sqrt{3}$, $(-4)^2 = 16$

Magnitude = $\sqrt{(4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}$

Unit vector $\vec{b} = \dfrac{1}{2\sqrt{6}}(\sqrt{3}-1, -\sqrt{3}-1, -4)$

To find the angle, we calculate the dot product of the unit vectors:

$\cos(\theta) = \vec{a} \cdot \vec{b}$

$\cos(\theta) = \dfrac{1}{\sqrt{6}} \cdot \dfrac{\sqrt{3}-1}{2\sqrt{6}} + \dfrac{1}{\sqrt{6}} \cdot \dfrac{-\sqrt{3}-1}{2\sqrt{6}} + \dfrac{-2}{\sqrt{6}} \cdot \dfrac{-4}{2\sqrt{6}}$

$\cos(\theta) = \dfrac{1}{12}[(\sqrt{3}-1) + (-\sqrt{3}-1) + 8]$

$\cos(\theta) = \dfrac{1}{12}[-2 + 8] = \dfrac{6}{12} = \dfrac{1}{2}$

Since $\cos(\theta) = \dfrac{1}{2}$, we have $\theta = \cos^{-1}(\dfrac{1}{2}) = 60°$

Therefore, the angle between the two lines is $60° = \dfrac{\pi}{3}$.