Share

Books Shortlist
Your shortlist is empty

# Solution for Find the Direction Cosines of the Lines, Connected by the Relations: L + M +N = 0 and 2lm + 2ln − Mn= 0. - CBSE (Commerce) Class 12 - Mathematics

ConceptDirection Cosines and Direction Ratios of a Line

#### Question

Find the direction cosines of the lines, connected by the relations: l + m +n = 0 and 2lm + 2ln − mn= 0.

#### Solution

$\text{ Given: }$

$l + m + n = 0 . . . \left( 1 \right)$

$2lm + 2\ln - nm = 0 . . . \left( 2 \right)$

$\text { From } \left( 1 \right), \text { we get }$

$l = - m - n$

$\text { Substituting } l = - m - n in \left( 2 \right), \text { we get }$

$2\left( - m - n \right)m + 2\left( - m - n \right)n - mn = 0$

$\Rightarrow - 2 m^2 - 2mn - 2mn - 2 n^2 - mn = 0$

$\Rightarrow 2 m^2 + 2 n^2 + 5mn = 0$

$\Rightarrow \left( m + 2n \right) \left( 2m + n \right) = 0$

$\Rightarrow m = - 2n, - \frac{n}{2}$

$\text { If } m = - 2n, \text { then from } \left( 1 \right), \text { we get } l = n .$

$\text { If } m = - \frac{n}{2}, \text { then from } \left( 1 \right), \text { we get } l = - \frac{n}{2} .$

$\text { Thus, the direction ratios of the two lines are proportional to } n, - 2n, n \text { and } - \frac{n}{2}, - \frac{n}{2}, n, \text { i . e } . 1, - 2, 1 \text { and } - \frac{1}{2}, - \frac{1}{2}, 1 .$

$\text { Hence, their direction cosines are }$

$\pm \frac{1}{\sqrt{6}}, \pm \frac{- 2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}$

$\pm \frac{- 1}{\sqrt{6}}, \pm \frac{- 1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [3]

Solution Find the Direction Cosines of the Lines, Connected by the Relations: L + M +N = 0 and 2lm + 2ln − Mn= 0. Concept: Direction Cosines and Direction Ratios of a Line.
S