If the direction cosines of the two lines sat | vedclass

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(D) Given equations for direction cosines $(l, m, n)$ are:$l+m+n=0 \Rightarrow m=-(l+n) \quad (i)$$2lm+2ln-mn=0 \quad (ii)$Substituting $m=-(l+n)$ int

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$60^{\circ}$

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(D) Given equations for direction cosines $(l, m, n)$ are:$l+m+n=0 \Rightarrow m=-(l+n) \quad (i)$$2lm+2ln-mn=0 \quad (ii)$Substituting $m=-(l+n)$ into $(ii)$:$2l(-(l+n)) + 2ln - (-(l+n))n = 0$$-2l^2 - 2ln + 2ln + ln + n^2 = 0$$-2l^2 + ln + n^2 = 0$$2l^2 - ln - n^2 = 0$$(2l+n)(l-n) = 0$This gives two cases:Case $1$: $l=n$. From $(i)$, $m=-(n+n)=-2n$. Thus, $(l, m, n) = (n, -2n, n)$, which gives direction ratios $(1, -2, 1)$.Case $2$: $2l=-n \Rightarrow l=-\frac{n}{2}$. From $(i)$, $m=-(-\frac{n}{2}+n) = -\frac{n}{2}$. Thus, $(l, m, n) = (-\frac{n}{2}, -\frac{n}{2}, n)$, which gives direction ratios $(1, 1, -2)$.Let the direction ratios be $\vec{a} = (1, -2, 1)$ and $\vec{b} = (1, 1, -2)$.The angle $	heta$ between the lines is given by:$\cos 	heta = \left| \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} ight| = \left| \frac{(1)(1) + (-2)(1) + (1)(-2)}{\sqrt{1^2+(-2)^2+1^2} \sqrt{1^2+1^2+(-2)^2}} ight|$$\cos 	heta = \left| \frac{1 - 2 - 2}{\sqrt{6} \sqrt{6}} ight| = \left| \frac{-3}{6} ight| = \frac{1}{2}$$	heta = \cos^{-1}\left(\frac{1}{2}ight) = 60^{\circ}$.

If the direction cosines of the two lines satisfy the equations $l+m+n=0$ and $2lm+2ln-mn=0$, then the acute angle between these lines is

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