If the lines with direction cosines (l, m, n) | vedclass

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(A) Eliminating $n$ from the given relations: $(fm + gl)(\frac{-al - bm}{c}) + hlm = 0$Or $ag(\frac{l}{m})^2 + (af + bg - ch)(\frac{l}{m}) + bf = 0 \d

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(A) Eliminating $n$ from the given relations: $(fm + gl)(\frac{-al - bm}{c}) + hlm = 0$Or $ag(\frac{l}{m})^2 + (af + bg - ch)(\frac{l}{m}) + bf = 0 \dots (1)$Let the roots of $(1)$ be $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$. Then $\frac{l_1}{m_1} \cdot \frac{l_2}{m_2} = \frac{bf}{ag} \Rightarrow \frac{l_1l_2}{f/a} = \frac{m_1m_2}{g/b} \dots (2)$Similarly,$\frac{m_1m_2}{g/b} = \frac{n_1n_2}{h/c} \dots (3)$From $(2)$ and $(3)$,$\frac{l_1l_2}{f/a} = \frac{m_1m_2}{g/b} = \frac{n_1n_2}{h/c} = \frac{l_1l_2 + m_1m_2 + n_1n_2}{(f/a) + (g/b) + (h/c)}$If the two lines are perpendicular,then $l_1l_2 + m_1m_2 + n_1n_2 = 0$Therefore,$\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$

If the lines with direction cosines $(l, m, n)$ satisfying $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular,then $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = .........$

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