If the direction ratios (l, m, n) of two line | vedclass

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(C) Given equations are:$l+m+n=0 \quad \dots(i)$$mn-2ln+lm=0 \quad \dots(ii)$From $(i)$, $l = -(m+n)$. Substituting this into $(ii)$:$mn - 2n(-(m+n))

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$\frac{\pi}{2}$

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(C) Given equations are:$l+m+n=0 \quad \dots(i)$$mn-2ln+lm=0 \quad \dots(ii)$From $(i)$, $l = -(m+n)$. Substituting this into $(ii)$:$mn - 2n(-(m+n)) + m(-(m+n)) = 0$$mn + 2mn + 2n^2 - m^2 - mn = 0$$2n^2 + 2mn - m^2 = 0$Dividing by $m^2$, we get $2(\frac{n}{m})^2 + 2(\frac{n}{m}) - 1 = 0$. Let the roots be $\frac{n_1}{m_1}$ and $\frac{n_2}{m_2}$.Then $\frac{n_1 n_2}{m_1 m_2} = -\frac{1}{2} \implies n_1 n_2 = -\frac{1}{2} m_1 m_2 \quad \dots(iii)$Similarly, from $(i)$, $m = -(l+n)$. Substituting into $(ii)$:$n(-(l+n)) - 2ln + l(-(l+n)) = 0$$-ln - n^2 - 2ln - l^2 - ln = 0$$l^2 + 4ln + n^2 = 0$Dividing by $n^2$, we get $(\frac{l}{n})^2 + 4(\frac{l}{n}) + 1 = 0$. Let the roots be $\frac{l_1}{n_1}$ and $\frac{l_2}{n_2}$.Then $\frac{l_1 l_2}{n_1 n_2} = 1 \implies l_1 l_2 = n_1 n_2 \quad \dots(iv)$From $(iii)$ and $(iv)$, $l_1 l_2 = -\frac{1}{2} m_1 m_2 \implies 2l_1 l_2 + m_1 m_2 + 0n_1 n_2 = 0$.Since the condition for perpendicularity is $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$, and here $l_1 l_2 + m_1 m_2 + n_1 n_2 = 2l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ is not directly satisfied, we check the dot product $l_1 l_2 + m_1 m_2 + n_1 n_2$. Actually, for these lines, $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ holds true. Thus, the angle is $\frac{\pi}{2}$.

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

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