Question

The angle between the two lines whose $d.c$ are given by equation $3l+m+5n=0$ and $6mn-2nl+5ml=0$

• A. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{6}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{4}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{1}{6}\right)$

Given lines are $3l+m+5n=0$

$m=-(3l+5n)$......(1)

and $6mn-2nl+5lm=0$.......(2)

Substitute m in equation (2) we obtain

$\begin{array}{c}\Rightarrow 6[-\left(3l+5n\right)]n-2nl+5l[-3l+5n]=0\\ \Rightarrow -18ln-30n^2-2nl-15l^2-25\ln =0\\ \Rightarrow -15l^2-45\ln -30n^2=0\\ \Rightarrow l^2+3\ln +2n^2=0\\ \Rightarrow l^2+2\ln +\ln +2n^2=0\\ \Rightarrow l\left(l+2n\right)+n\left(l+2n\right)=0\\ \Rightarrow \left(l+n\right)\left(l+2n\right)=0\\ \therefore l=-nandl=-2n\\ \left(l/-1\right)=\left(n/1\right)and\left(l/-2\right)=\left(n/1\right)......\left(3\right)\\ substitutelinequationweget\\ m=-\left(3l+5n\right)\\ m=-2nandm=n\\ \left(m/-2\right)=\left(n/1\right)and\left(m/1\right)=\left(n/1\right).....\left(4\right)\\ From\left(3\right)and\left(4\right)weget\\ \left(l/-1\right)=\left(m/-2\right)=\left(n/1\right)and\left(l/-2\right)=\left(m/1\right)=\left(n/1\right)\\ l:m:n=-1:-2:1andl:m:n=-2:1:1\\ i.e.D.r^{\prime }s\left(-1,-2,1\right)and\left(-2,1,1\right)\\ Anglebetweenthelineswhosedirection\cos inesare\\ \cos \theta =\frac{-1\times \left(-2\right)+\left(-2\right)\times 1+1\times 1}{\sqrt{{\left(-1\right)}^2+{\left(-2\right)}^2+1^2}\sqrt{{\left(-2\right)}^2+1^2+1^2}}\\ \cos \theta =\frac{1}{\sqrt{6}\sqrt{6}}\\ \cos \theta =\frac{1}{6}\\ \therefore \theta ={\cos }^{-1}\left(\frac{1}{6}\right)\end{array}$