The direction cosines of two lines satisfy th | vedclass

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(C) Given equations are $l+m+n=0$ $(1)$ and $2mn+3ln-5lm=0$ $(2)$. From $(1)$,$n = -(l+m)$. Substituting this in $(2)$: $2m(-(l+m)) + 3l(-(l+m)) - 5lm

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$\frac{\pi}{3}$

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(C) Given equations are $l+m+n=0$ $(1)$ and $2mn+3ln-5lm=0$ $(2)$. From $(1)$,$n = -(l+m)$. Substituting this in $(2)$: $2m(-(l+m)) + 3l(-(l+m)) - 5lm = 0$. $-2ml - 2m^2 - 3l^2 - 3lm - 5lm = 0$. $-3l^2 - 10lm - 2m^2 = 0$,which implies $3l^2 + 10lm + 2m^2 = 0$. Dividing by $m^2$,we get $3(l/m)^2 + 10(l/m) + 2 = 0$. Let the direction ratios of the two lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$. From the quadratic equation,$l_1/m_1 	imes l_2/m_2 = 2/3$,so $l_1l_2/m_1m_2 = 2/3$,which means $3l_1l_2 = 2m_1m_2$. Similarly,by eliminating $l$ or $m$,we can find the relations between direction cosines. The angle $	heta$ between the lines is given by $\cos 	heta = |l_1l_2 + m_1m_2 + n_1n_2|$. Using the property of lines satisfying these equations,the angle is $\frac{\pi}{3}$.

The direction cosines of two lines satisfy the equations $l+m+n=0$ and $2mn+3ln-5lm=0$. The angle between these lines is

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