Question

The angle between the lines whose direction cosines are connected by the relations $l+m+n=0$ and $2lm+2nl-mn=0$ is

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

Answer 1

The correct option is A $\frac{\pi }{2}$

$l+m+n=0$ . . . eq.1

$2lm+2nl-mn=0$ . . . eq.2

From eq.1, $n=-(l+m)$

eq.2 implies,

$2lm+2l[-(l+m\left)\right]-m[-(l+m\left)\right]=0\Rightarrow 2lm-(2l^2+2lm)+ml+m^2=0\Rightarrow -2l^2+ml+m^2=0\Rightarrow 2l^2-ml-m^2=0\Rightarrow 2l(l-m)+m(l-m)=0\Rightarrow (l-m)(2l+m)=0$

$\therefore$ Either $l=m$ or, $l=\frac{-m}{2}$

If $l=m$ then $n=-2m$

f $l=\frac{-m}{2}$ then $n=-(\frac{-m}{2}+m)=\frac{-m}{2}$

$\therefore$ The direction ratios of the straight lines are $(m,m-2m)$ & $(-m/2,-2m,-m/2)$

i.e. multiples of $(1,1,2)$ & $(-1/2,1,-1/2)$

If the angle between the lines is $\theta$

then $\cos \theta =\frac{1.\left(\frac{-1}{2}\right)+1.1+(-2).\left(\frac{-1}{2}\right)}{\sqrt{1+1+4}.\sqrt{\frac{1}{4}+1+\frac{1}{4}}}=\frac{\frac{-1}{2}+1+1}{\sqrt{4+1+1}.\sqrt{\frac{1}{2}+1}}=\frac{3/2}{\sqrt{6.\frac{3}{2}}}=\frac{3/2}{3}=\frac{1}{2}\Rightarrow \theta ={\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}.$

$\therefore$ The angles between the lines is $\frac{\pi }{3}$