The acute angle between the two lines whose direction ratios are connected by $l+m-n=0and{l}^2+m^2-n^2=0$ is

Solution:

Suppose that $n=1$

Hence the equations become $l+m=1$and $l^2+m^2=1$

Substitute $m=1-l$ in the equation  $l^2+m^2=1$and then solve for $l$

It implies that $l=1,0$ and $m=0,1$

Therefore, the direction ratios are $〈1,0,1〉,〈0,1,1〉$

If $\theta$ is the acute angle between two lines having direction ratios $〈a_1,b_1,c_1〉$and $〈a_2,b_2,c_2〉$

 then $\cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\cdot \sqrt{{a_2}^2+{b_2}^2+{c_2}^2}}\right|$

Suppose that $\theta$ be the angle between the lines whose direction ratios are $〈1,0,1〉,〈0,1,1〉$

Then $\cos \theta =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

Therefore, the measure of angle $\theta$ is $\boxed{\frac{\pi }{3}}$