Question

Find the angle between the two lines whose direction cosines are given by equations $l+m+n=0$ and $l^2+m^2-n^2=0$

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

Answer 1

The correct option is B $\frac{\pi }{3}$

Given that $l+m+n=0$.....(1)

$\Rightarrow l+m=-n$

$\Rightarrow -\left(l+m\right)=n$

$and{l}^2+m^2-n^2=\mathrm{0......}\left(2\right)$

$Letussubstitutefor{}^{\prime }{n}^{\prime }inequation\left(2\right)weget$

$\Rightarrow l^2+m^2-l^2-m^2-2ml=0$

$or2ml=0$

$eitherl=0orm=0$

$letusputm=0inequation\left(1\right)$

$ifm=0thenl=-n$

$directionratios\left(l,m,n\right)=\left(1,0,-1\right)$

$letusputl=0wegetm=-n$

$directionratios\left(l,m,n\right)=\left(0,1,-1\right)$

$letusfindout{\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2$

${\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2=\left(1,0,-1\right)\cdot \left(0,1,-1\right)$

$=0+0+1=1$

$\left|{\overrightarrow{b}}_1\right|=\sqrt{0^2+1^2+{\left(-1\right)}^2}=\sqrt{2}$

$\left|{\overrightarrow{b}}_2\right|=\sqrt{0^2+1^2+{\left(-1\right)}^2}=\sqrt{2}$

$Nowsubstitutingtheabovevaluein$

$\cos \theta =\frac{\overrightarrow{b_1}\cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

$\theta =\frac{\pi }{3}$