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# Solution for Find the Angle Between the Lines Whose Direction Cosines Are Given by the Equations 2l − M + 2n = 0 And Mn + Nl + Lm = 0 - CBSE (Science) Class 12 - Mathematics

ConceptDirection Cosines and Direction Ratios of a Line

#### Question

Find the angle between the lines whose direction cosines are given by the equations

2l − m + 2n = 0 and mn + nl + lm = 0

#### Solution

$\left( ii \right) \text{ Given } :$

$2l - m + 2n = 0 . . . (1)$

$mn + nl + lm = 0 . . . (2)$

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

$m = 2l + 2n$

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

$\left( 2l + 2n \right)n + nl + l\left( 2l + 2n \right) = 0$

$\Rightarrow 2\ln + 2 n^2 + nl + 2 l^2 + 2\ln = 0$

$\Rightarrow 2 l^2 + 5ln + 2 n^2 = 0$

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

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

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

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

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

$\text{ Vectors parallel these lines are }$

$\vec{a} = - 2 \hat{i} - \hat{2j} + \hat{k}$

$\vec{b} = - \frac{1}{2} \hat{i} + \hat{j} + \hat{k}$

$\text{ If } \theta \text{ is the angle between the lines, then } \theta \text{ is also the angle between } \vec{a} \text { and } \vec{b .}$

$\text{ Now },$

$\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|}$

$= \frac{1 - 2 + 1}{\sqrt{4 + 4 + 1} \sqrt{ 1/4 + 1 + 1}}$

$= 0$

$\Rightarrow \theta = \frac{\pi}{2}$

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Solution Find the Angle Between the Lines Whose Direction Cosines Are Given by the Equations 2l − M + 2n = 0 And Mn + Nl + Lm = 0 Concept: Direction Cosines and Direction Ratios of a Line.
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