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# Find the Angle Between the Lines Whose Direction Cosines Are Given by the Equations L + 2m + 3n = 0 and 3lm − 4ln + Mn = 0 - CBSE (Arts) 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

l + 2m + 3n = 0 and 3lm − 4ln + mn = 0

#### Solution

 \text{ Given } :

$l + 2m + 3n = 0 . . . (1)$

$3lm - 4\ln + mn = 0 . . . (2)$

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

$l = - 2m - 3n$

$\text { Substituting } l = - 2m - 3n \text { in } \left( 2 \right), \text { we get }$

$3\left( - 2m - 3n \right)m - 4\left( - 2m - 3n \right)n + mn = 0$

$\Rightarrow - 6 m^2 - 9mn + 8mn + 12 n^2 + mn = 0$

$\Rightarrow 12 n^2 - 6 m^2 = 0$

$\Rightarrow m^2 = 2 n^2$

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

$\text{ If } m = \sqrt{2}n,\text { then by substituting } m = \sqrt{2}n \text { in } \left( 1 \right), \text { we get } l = n\left( - 2\sqrt{2} - 3 \right) .$

$\text { If } m = - \sqrt{2} n, \text { then by substituting }m = - \sqrt{2} n \text { in } \left( 1 \right), \text { we get } l = n\left( 2\sqrt{2} - 3 \right) .$

$\text { Thus, the direction ratios of the two lines are proportional to } n\left( - 2\sqrt{2} - 3 \right), \sqrt{2}n, n \text { and } n\left( 2\sqrt{2} - 3 \right), - \sqrt{2} n, n \text { or } \left( - 2\sqrt{2} - 3 \right), \sqrt{2} , 1 \text{ and } \left( - 2\sqrt{2} - 3 \right), - \sqrt{2}, 1 .$

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

$\vec{a} = \left( - 2\sqrt{2} - 3 \right) \hat{i} + \sqrt{2} \hat{j} + \hat{k}$

$\vec{b} = \left( 2\sqrt{2} - 3 \right) \hat{i} - \sqrt{2} \hat{ j} + \hat{k}$

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

$\text{ Now},$

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

$= \frac{\left[ \left( - 2\sqrt{2} - 3 \right) \hat{i} + \sqrt{2} \hat{j} + \hat{k} \right] . \left[ \left( 2\sqrt{2} - 3 \right) \hat{i} - \sqrt{2} \hat{j} + \hat{k} \right]}{\sqrt{8 + 9 + 12\sqrt{2} + 2 + 1} \sqrt{8 + 9 - 12\sqrt{2} + 2 + 1}}$

$= \frac{- \left( 8 - 9 \right) - 2 + 1}{\sqrt{20 + 12\sqrt{2}} \sqrt{20 - 12\sqrt{2}}}$

$= \frac{0}{\sqrt{20 + 12\sqrt{2}} \sqrt{20 - 12\sqrt{2}}}$

$= 0$

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Solution for Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (2018 to Current)
Chapter 27: Direction Cosines and Direction Ratios
Ex.27.1 | Q: 16.3 | Page no. 23

#### Video TutorialsVIEW ALL [3]

Solution Find the Angle Between the Lines Whose Direction Cosines Are Given by the Equations L + 2m + 3n = 0 and 3lm − 4ln + Mn = 0 Concept: Direction Cosines and Direction Ratios of a Line.
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