If the direction cosines between two lines are given by the relations $a l + b m + c n = 0$ and $h l m + f m n + g n l = 0$ , then the two lines are parallel. if $\sqrt{a f} \pm \sqrt{b g} \pm \sqrt{c h} =$

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Detailed Solution

Given al + bm + cn = 0 and fmn + gnl + hlm = 0 Eliminating n, we get
$\begin{aligned} f m (- \frac{a l + b m}{c}) + g l (- \frac{a l + b m}{c}) + h l m = 0 \\ \Rightarrow - a f m - b f m^{2} - a g l^{2} - b g l m + c h l m = 0 \\ \Rightarrow a g {(\frac{1}{m})}^{2} + (a f + b g - c h) (\frac{l}{m}) + b f = 0 \end{aligned}$
$Let its roots are \frac{l_{1}}{m_{1}} and \frac{l_{2}}{m_{2}} . N o w,$
$\begin{array}{l} \frac{l_{1}}{m_{1}} \cdot \frac{l_{2}}{m_{2}} = \frac{b f}{a g} \\ \Rightarrow \frac{l_{2} l_{2}}{b f} = \frac{m_{1} m_{2}}{a g} \\ \Rightarrow \frac{l_{2} l_{2}}{b f} = \frac{m_{1} m_{2}}{a g} = \frac{c n_{1} n_{2}}{h} \\ \Rightarrow \frac{l_{2} l_{2}}{(f l a)} = \frac{m_{1} m_{2}}{(g / b)} = \frac{n_{1} n_{2}}{(h / c)} \end{array}$
If the lines are parallel, so l₁/l₂ = m₁/m₂= n₁/n₂ So, the roots of the Eq. (i) have equal roots, i.e. D = 0
$\begin{array}{ll} \Rightarrow (a f + b g - c h)^{2} = 4 a b f g \\ \Rightarrow (a f + b g - c h) = \pm 2 \sqrt{a f} \sqrt{b g} \\ \Rightarrow (a f + b g \pm 2 \sqrt{a f} \sqrt{b g}) = c h \\ \Rightarrow (\sqrt{a f} \pm \sqrt{b g})^{2} = (\sqrt{c h})^{2} \\ \Rightarrow (\sqrt{a f} \pm \sqrt{b g}) = \pm (\sqrt{c h}) \\ \Rightarrow (\sqrt{a f} \pm \sqrt{b g} + \sqrt{c h}) = 0 \end{array}$