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प्रश्न
Find the direction cosines of the lines, connected by the relations: l + m +n = 0 and 2lm + 2ln − mn= 0.
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उत्तर
\[\text{ Given: } \]
\[l + m + n = 0 . . . \left( 1 \right)\]
\[2lm + 2\ln - nm = 0 . . . \left( 2 \right)\]
\[\text { From } \left( 1 \right), \text { we get } \]
\[l = - m - n\]
\[\text { Substituting } l = - m - n in \left( 2 \right), \text { we get }\]
\[2\left( - m - n \right)m + 2\left( - m - n \right)n - mn = 0\]
\[ \Rightarrow - 2 m^2 - 2mn - 2mn - 2 n^2 - mn = 0\]
\[ \Rightarrow 2 m^2 + 2 n^2 + 5mn = 0\]
\[ \Rightarrow \left( m + 2n \right) \left( 2m + n \right) = 0 \]
\[ \Rightarrow m = - 2n, - \frac{n}{2}\]
\[\text { If } m = - 2n, \text { then from } \left( 1 \right), \text { we get } l = n . \]
\[\text { If } m = - \frac{n}{2}, \text { then from } \left( 1 \right), \text { we get } l = - \frac{n}{2} . \]
\[\text { Thus, the direction ratios of the two lines are proportional to } n, - 2n, n \text { and } - \frac{n}{2}, - \frac{n}{2}, n, \text { i . e } . 1, - 2, 1 \text { and } - \frac{1}{2}, - \frac{1}{2}, 1 . \]
\[\text { Hence, their direction cosines are } \]
\[ \pm \frac{1}{\sqrt{6}}, \pm \frac{- 2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}} \]
\[ \pm \frac{- 1}{\sqrt{6}}, \pm \frac{- 1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}\]
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