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# Solution for Find the Angle Between the Lines Whose Direction Cosines Are Given by the Equations (I) L + M + N = 0 and L2 + M2 − N2 = 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
(i) m + n = 0 and l2 + m2 − n2 = 0

#### Solution

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

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

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

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

$m = - l - n$

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

$l^2 + \left( - l - n \right)^2 - n^2$

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

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

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

$\Rightarrow l = 0 , l = - n$

$\text{ If } l = 0, \text{ then by substituting } l = 0 \text { in } \left( 1 \right), \text { we get } m = - n .$

$\text{ If } l = - n, \text { then by substituting } l = - n \text { in }\left( 1 \right), \text { we get } m = 0 .$

$\text{ Thus, the direction ratios of the two lines are proportional to } 0, - n, \text { and } - n, 0, n \text{ or } 0, - 1, 1 \text { and } - 1, 0, 1 .$

$\text{ Vectors parallel to these lines are }$

$\vec{a} = 0 \hat{i} - \hat{j} + \hat{k}$

$\vec{b} = - \hat{i} + 0 \hat{j} + \hat{k}$

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

$\text { Now },$

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

$= \frac{1}{\sqrt{0 + 1 + 1} \sqrt{1 + 0 + 1}}$

$= \frac{1}{2}$

$\Rightarrow \theta = \frac{\pi}{3}$

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#### Video TutorialsVIEW ALL [3]

Solution Find the Angle Between the Lines Whose Direction Cosines Are Given by the Equations (I) L + M + N = 0 and L2 + M2 − N2 = 0 Concept: Direction Cosines and Direction Ratios of a Line.
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