If (l_1, m_1, n_1) and (l_2, m_2, n_2) are th | vedclass

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(B) Given equations are $2l-m+3n=0$ and $l^2+mn-6n^2=0$. From the first equation,$m=2l+3n$. Substituting this into the second equation: $l^2+(2l+3n)n-

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$\frac{2 \sqrt{3}}{\sqrt{19}}$

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(B) Given equations are $2l-m+3n=0$ and $l^2+mn-6n^2=0$. From the first equation,$m=2l+3n$. Substituting this into the second equation: $l^2+(2l+3n)n-6n^2=0$. $l^2+2ln+3n^2-6n^2=0 \Rightarrow l^2+2ln-3n^2=0$. Factoring gives $(l+3n)(l-n)=0$,so $l=n$ or $l=-3n$. Case $1$: If $l=n$,then $m=2(n)+3n=5n$. The direction ratios are $(n, 5n, n)$,or $(1, 5, 1)$. The direction cosines are $(\frac{1}{\sqrt{1^2+5^2+1^2}}, \frac{5}{\sqrt{27}}, \frac{1}{\sqrt{27}}) = (\frac{1}{3\sqrt{3}}, \frac{5}{3\sqrt{3}}, \frac{1}{3\sqrt{3}})$. Case $2$: If $l=-3n$,then $m=2(-3n)+3n=-3n$. The direction ratios are $(-3n, -3n, n)$,or $(-3, -3, 1)$. The direction cosines are $(\frac{-3}{\sqrt{(-3)^2+(-3)^2+1^2}}, \frac{-3}{\sqrt{19}}, \frac{1}{\sqrt{19}}) = (\frac{-3}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{1}{\sqrt{19}})$. Thus,$l_1=\frac{1}{3\sqrt{3}}, m_1=\frac{5}{3\sqrt{3}}$ and $l_2=\frac{-3}{\sqrt{19}}, m_2=\frac{-3}{\sqrt{19}}$. Calculating $|l_1 l_2|+|m_1 m_2| = |(\frac{1}{3\sqrt{3}})(\frac{-3}{\sqrt{19}})| + |(\frac{5}{3\sqrt{3}})(\frac{-3}{\sqrt{19}})| = |\frac{-1}{\sqrt{3}\sqrt{19}}| + |\frac{-5}{\sqrt{3}\sqrt{19}}| = \frac{1}{\sqrt{57}} + \frac{5}{\sqrt{57}} = \frac{6}{\sqrt{57}} = \frac{6}{\sqrt{3}\sqrt{19}} = \frac{2\sqrt{3}}{\sqrt{19}}$.

If $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines satisfying the relations $l^2+mn-6n^2=0$ and $2l-m+3n=0$,then $|l_1 l_2|+|m_1 m_2|=$

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