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(D) Given the equations for the direction ratios $(l, m, n)$:$l+m-n=0$ $(i)$$l^2+m^2-n^2=0$ (ii)From equation $(i)$, we have $l = n-m$.Substituting th

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$\frac{\pi}{3}$

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(D) Given the equations for the direction ratios $(l, m, n)$:$l+m-n=0$ $(i)$$l^2+m^2-n^2=0$ (ii)From equation $(i)$, we have $l = n-m$.Substituting this into equation (ii):$(n-m)^2 + m^2 - n^2 = 0$$n^2 + m^2 - 2nm + m^2 - n^2 = 0$$2m^2 - 2nm = 0$$2m(m-n) = 0$This gives two cases: $m=0$ or $m=n$.Case $1$: If $m=0$, then $l=n$. The direction ratios are $(1, 0, 1)$.Case $2$: If $m=n$, then $l=0$. The direction ratios are $(0, 1, 1)$.Let the two lines have direction ratios $\vec{a} = (1, 0, 1)$ and $\vec{b} = (0, 1, 1)$.The cosine of the angle $	heta$ between them is given by:$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(0) + (0)(1) + (1)(1)|}{\sqrt{1^2+0^2+1^2} \sqrt{0^2+1^2+1^2}}$$\cos 	heta = \frac{1}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$Since $\cos 	heta = \frac{1}{2}$, the acute angle is $	heta = \frac{\pi}{3}$.

The acute angle between the two lines whose direction ratios $(l, m, n)$ satisfy the equations $l+m-n=0$ and $l^2+m^2-n^2=0$ is

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