Two lines whose direction cosines are given b | vedclass

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(A) Let the direction cosines of the lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$

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$\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$

Vedclass · Answer

(A) Let the direction cosines of the lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$.From the first equation,$l = -\frac{bm + cn}{a}$. Substituting this into the second equation:$fmn + gn(-\frac{bm + cn}{a}) + hm(-\frac{bm + cn}{a}) = 0$$afmn - bgnm - cgn^2 - bhm^2 - chmn = 0$$bhm^2 + (ch + bg - af)mn + cgn^2 = 0$Dividing by $n^2$,we get $bh(\frac{m}{n})^2 + (ch + bg - af)(\frac{m}{n}) + cg = 0$.The product of the roots is $\frac{m_1 m_2}{n_1 n_2} = \frac{cg}{bh}$.Similarly,by eliminating $m$,we get $ah(\frac{l}{n})^2 + (ch + af - bg)(\frac{l}{n}) + cf = 0$,so $\frac{l_1 l_2}{n_1 n_2} = \frac{cf}{ah}$.Since the lines are perpendicular,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.Dividing by $n_1 n_2$,we get $\frac{l_1 l_2}{n_1 n_2} + \frac{m_1 m_2}{n_1 n_2} + 1 = 0$.Substituting the products: $\frac{cf}{ah} + \frac{cg}{bh} + 1 = 0$.Dividing by $c$,we get $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

Two lines whose direction cosines are given by $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular to each other if .........

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