Question

Find the angles between the lines whose direction cosines are given by the equations.

$l+m+n=0$

$l^2+m^2+n^2=0$

Answer 1

According to question,

$l+m+n=0,l^2+m^2+n^2=0$

so,

Angle between two lines is $\cos \theta =\frac{\overline{b_1}.{\overline{b}}_2}{\left|b_1\right|\left|b_2\right|}$

$\begin{array}{c}\cos \theta =\frac{\overline{b_1}.{\overline{b}}_2}{\left|b_1\right|\left|b_2\right|}\\ step1:l+m+n=0----\left(1\right)\\ l+m=-n\\ n=-(l+m)\\ and\\ l^2+m^2+n^2=0-------\left(2\right)\\ l^2+m^2+(-(l+m){)}^2=0\\ l^2+m^2+(-(l^2+m^2+2lm\left)\right)=0\\ l^2+m^2-l^2-m^2-2lm=0\\ \therefore lm=0,\\ i.eeitherl=0,m=0\\ letusputm=0inequ----\left(1\right)\end{array}$

$\begin{array}{c}ifm=0thenl=-n\left(d{r}^{\prime }sisproportional\right)\\ directionratio(l,m,n)=(1,0,-1)----(b_1\\ putl=0,thenwegetm=-n\\ d{r}^{\prime }s(l,m,n).(0,1,-1)---(b_2\\ steps2:\\ b_1{b}_2=(1,0,-1).(0,1,-1)=1\\ \left|b_1\right|=\sqrt{1+01}=\sqrt{2},\left|b_2\right|=\sqrt{2}\\ steps3:\\ \cos \theta =\frac{\overline{b_1}\overline{b_2}}{\left|b_1\right|\left|b_2\right|}=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\\ \therefore \theta =\frac{\pi }{3}\end{array}$