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  • The angle between the straight lines, whose direction cosines are given by the equations <img alt="2 l+2 m-n=0$ and $m n+n l+l m=0" src="https://entrancecorner.codecogs.com/gif.latex?2%20l&plus;2%20m-n%3D0%24%20and%20%24m%20n&plus;n%20l&plus;l

The angle between the straight lines, whose direction cosines are given by the equations 2 l+2 m-n=0$ and $m n+n l+l m=0, is :
Option: 1 \pi-\cos ^{-1}\left(\frac{4}{9}\right)
Option: 2 \frac{\pi}{2}
Option: 3 \cos ^{-1}\left(\frac{8}{9}\right)
Option: 4 \frac{\pi}{3}

Answers (1)

best_answer

2l+2m-n= 0\\

\Rightarrow n= 2l+2m\\            ...............(1)

mn+nl+lm= 0\\

\Rightarrow n\left ( m+l \right )+lm= 0\\

\Rightarrow 2\left ( l+m \right )\left ( l+m \right )+lm= 0\\

\Rightarrow 2l^{2}+4ml+2m^{2}+lm= 0\\

\Rightarrow 2l\left ( l+2m \right )+m\left ( l+2m \right )= 0\\

\Rightarrow \left ( 2l+m \right )\left ( l+2m \right )= 0\\

\Rightarrow m= -2l\: or\: l= -2m\\               ..........(2)

From (1) and (2)

n= 2l+2m\\

   = 2l-4l= -2l\: or\: -4m+2m= -2m\\

So m= -2l= n\:\: OR\: \: l= -2m= n\\

Also l^{2}+m^{2}+n^{2}= 1\\

\Rightarrow l^{2}+4l^{2}+4l^{2}-1\Rightarrow l= \pm \frac{1}{3}\\

OR\: \: 4m^{2}+m^{2}+4m^{2}= 1\Rightarrow m= \frac{1}{3}\\

So L_{1}:\pm \left ( \frac{1}{3} ,\frac{-2}{3},\frac{-2}{3}\right )\: L_{2}:\pm \left ( \frac{-2}{3},\frac{1}{3}, \frac{-2}{3}\right )\\

Angle \cos\theta= \left | l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2} \right |\\

                    = \left | \frac{-2}{9}-\frac{2}{9}+\frac{4}{9} \right |= 0\\

                    \Rightarrow \theta= 90^{\circ}= \frac{\pi}{2}

Posted by

Kuldeep Maurya

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