Prove that the lines whose direction cosines are given by the equations `l+m+n=0 and 3lm-5mn+2nl=0` are mutually perpendicular.

Explanation

Let `l_(1), m_(1), n_(1), and l_(2), m_(2), n_(2)` be the direction cosines of the two lines satisfying the equations
`l+ m+ n=0 and 3lm- 5mn +2nl +0`
`rArr 3lm + 5m (l+m) - 2l (l+m)=0`
`rArr 3lm + 5lm +5m^(2) - 2l^(2) -2lm=0`
`rArr 2l^(2)-6lm-5m^(2)=0`
rArr `2(l/m)^(2)-6(l/m)-5=0`
`l_(1)/m_(1) +l_(2)/m_(2)=3, l_(1)/m_(1)cdot l_(2)/m_(2)=-5/2`
` rArr (l_(1)l_(2))/(-5)=(m_(1)m_(2))/2 " ".... (i)`
Again, eliminating m between above equations
we get,
`-3l(l+n)+5n(l+n)+2nl =0`
`rArr 3l^(2) - 4ln -5n^(2) =0`
`rArr 3(l/n)^(2)-4l/n-5=0`
`rArr (l_(1)l_(2))/(n_(1)n_(2))=-5/3`
`rArr (l_(1)l_(2))/(-5)=(n_(1)n_(2))/3" "...(ii)`
From (i) and (ii) it follows that
`(l_(1)l_(2))/(-5)=(m_(1)m_(2))/2=(n_(1)n_(2))/3=(l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2))/(-5+2+3)`
`rArr l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=0`
`rArr` the two straight lines are perpendicular.