When calculating the angle between two lines in 3D, the dot product and the magnitudes of the direction vectors are crucial. Remember to use the formula \( \cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} \), where \( \vec{d_1} \cdot \vec{d_2} \) is the dot product and \( |\vec{d_1}| \) and \( |\vec{d_2}| \) are the magnitudes of the direction vectors. The angle is then found using the inverse cosine function. Pay close attention to the calculation of magnitudes, as the square roots can sometimes lead to tricky simplifications.
The angle \( \theta \) between two lines with direction ratios \( \vec{d_1} = (1, 1, -2) \) and \( \vec{d_2} = (\sqrt{3} - 1, -\sqrt{3} - 1, -4) \) is calculated using the formula:
\[\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}||\vec{d_2}|}.\]
The dot product \( \vec{d_1} \cdot \vec{d_2} \) is:
\[\vec{d_1} \cdot \vec{d_2} = (1)(\sqrt{3} - 1) + (1)(-\sqrt{3} - 1) + (-2)(-4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6.\]
The magnitudes are:
\[|\vec{d_1}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6},\]
\[|\vec{d_2}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + (-4)^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{24} = 2\sqrt{6}.\]
Substituting these values:
\[\cos \theta = \frac{6}{\sqrt{6} \times 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2}.\]
Thus,
\[\theta = \cos^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{3}.\]
The resulting angle is: \(\frac{\pi}{3}\).