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The angle between the lines whose $d.c.$ are given by the equations $3l+m+5n=0$ and $6mn-2nl+5lm=0$ is?

$\begin{array}{1 1} (a)\:\:0\:\:\:\:\qquad\:\:(b)\:\:\large\frac{\pi}{2}\:\:\:\:\qquad\:\:(c)\:\:cos^{-1}\large\frac{1}{3}\:\:\:\:\qquad\:\:(d)\:\:cos^{-1}\large\frac{1}{6} \end{array} $

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1 Answer

  • Angle between the two lines with d.r.$\overrightarrow b_1$ and $\overrightarrow b_2$ is given by $cos\theta=\large\frac{\overrightarrow b_1.\overrightarrow b_2}{|\overrightarrow b_1|\:|\overrightarrow b_2|}$
Given equations are $3l+m+5n=0$........(i) and $6mn-2nl+5lm=0$...........(ii)
$\Rightarrow\:m=-5n-3l$, Substituting this in (ii) we get
$\Rightarrow\:(2n+l)(n+l)=0$ or $l=-2n\:\:or\:\:l=-n$
If $n=1,$ then $(l,m,n)=(-2,1,1)\:\:or\:\:(1,2,-1)$
$\therefore\:d.r.$ of the lines $ (l,m,n)$ can be taken as $ (-2,1,1) \:\: and\:\:(1,2,-1)$
$\therefore \:$ Angle between the lines is $cos\theta=\large\frac{(-2,1,1).(1,2,-1)}{\sqrt {6}.\sqrt 6}=\large\frac{-2+2-1}{6}$
$\Rightarrow\:\theta=cos ^{-1}\large\frac{1}{6}$
answered Jan 1, 2014 by rvidyagovindarajan_1

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