The acute angle between the two lines whose d | vedclass

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

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$\frac{\pi}{3}$

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(D) Given the equations for direction ratios $(l, m, n)$:$l+m-n=0 \quad (i)$$l^2+m^2-n^2=0 \quad (ii)$From equation $(i)$,we have $l = n-m$.Substituting this into equation $(ii)$:$(n-m)^2 + m^2 - n^2 = 0$$n^2 + m^2 - 2nm + m^2 - n^2 = 0$$2m^2 - 2nm = 0$$2m(m-n) = 0$This gives two cases:Case $1$: $m=0$. From $(i)$,$l=n$. Thus,the direction ratios are $(1, 0, 1)$.Case $2$: $m=n$. From $(i)$,$l=0$. Thus,the direction ratios are $(0, 1, 1)$.Let $	heta$ be the acute angle between the two lines with direction ratios $\vec{a} = (1, 0, 1)$ and $\vec{b} = (0, 1, 1)$.$\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} \cdot \sqrt{2}} = \frac{1}{2}$Since $\cos 	heta = \frac{1}{2}$,we have $	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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