The angle between two lines whose direction ratios are proportional to 1, 1,−2 and (√3−1),(−√3−1),−4 is:

The correct option is π/3.

To find the angle between two lines whose direction ratios are given, we use the angle formula for direction ratios.

Let the direction ratios of the first line be $a_1,b_1,c_1$ and the direction ratios of the second line be $a_2,b_2,c_2$.

Here, the direction ratios of the first line are:
$a_1=1$, $b_1=1$, $c_1=-2$

The direction ratios of the second line are:
$a_2=\sqrt{3}-1$, $b_2=-\sqrt{3}-1$, $c_2=-4$

The angle $\theta$ between two lines with direction ratios is given by the formula:
$\cos \left(\theta \right)=\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}·\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}}$

First, let us calculate the numerator, which is the dot product of the direction ratios:
$a_1{a}_2+b_1{b}_2+c_1{c}_2=1·(\sqrt{3}-1)+1·(-\sqrt{3}-1)+(-2)·(-4)$
$=\sqrt{3}-1-\sqrt{3}-1+8$
$=-2+8=6$

Next, let us calculate the magnitude of the first direction ratio vector:
$\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}=\sqrt{1^2+1^2+{(-2)}^2}=\sqrt{1+1+4}=\sqrt{6}$

Now, let us calculate the magnitude of the second direction ratio vector:
$\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}=\sqrt{{(\sqrt{3}-1)}^2+{(-\sqrt{3}-1)}^2+{(-4)}^2}$

Let us expand the terms inside the square root:
${(\sqrt{3}-1)}^2=3-2\sqrt{3}+1=4-2\sqrt{3}$
${(-\sqrt{3}-1)}^2={(\sqrt{3}+1)}^2=3+2\sqrt{3}+1=4+2\sqrt{3}$
${(-4)}^2=16$

Adding these together:
${a_2}^2+{b_2}^2+{c_2}^2=(4-2\sqrt{3})+(4+2\sqrt{3})+16$
$=4+4+16=24$

Taking the square root, we get:
$\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}=\sqrt{24}=2\sqrt{6}$

Now, substituting these values back into the cosine formula:
$\cos \left(\theta \right)=\frac{6}{\sqrt{6}·2\sqrt{6}}$
$\cos \left(\theta \right)=\frac{6}{2·6}=\frac{6}{12}=\frac{1}{2}$

Since $\cos \left(\theta \right)=\frac{1}{2}$, the angle is:
$\theta ={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$