Question

If the $d.c^{\prime }s$ of two lines are connected by the equations $l+m+n=0,l^2+m^2-n^2=0$, then angle between the lines is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{3}$

Given equation are,

$\begin{array}{c}l+m+n=0----------\left(1\right)\\ l^2+m^2+n^2=0---------\left(2\right)\\ fromequ\left(1\right),l=-(m+n)\\ valueofl.......\\ substitutingin\left(2\right)\\ (m+n{)}^2+m^2-n^2=0\\ \Rightarrow m^2+n^2+2mn+m^2-n^2=0\\ \Rightarrow 2m^2+2mn=0\\ \Rightarrow 2m(m+n)=0\\ \Rightarrow m=0and(m+n)=0\\ Now,\\ m=0,substitutingin\left(1\right)\\ l+m+n=0\\ l+n=0,\Rightarrow l=-n\\ Dr.softhefirstlineare(1,0,-1)\\ and,\\ m+n=0\Rightarrow m=-n\\ substitutingin\left(1\right)\\ l+m+n=0\\ l=0\\ D^{\prime }rsofthesecondlineare(0,1,-1)\end{array}$

$\begin{array}{c}Lets\theta betheanglebetweentheline,\\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\begin{array}{c}\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}\\ =\frac{0+0+1}{\sqrt{2}.\sqrt{2}}=\frac{1}{2}\\ \cos \theta =\frac{1}{2}\\ \therefore \theta =\frac{\pi }{3}\end{array}}\end{array}$

so that the correct option is A.