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

Explanation

Eliminating `n` form both the euations we have
`l^(2)+m^(2)-(l-m)^(2)=0`
`implies l^(2)+m^(2)-l^(2)-m^(2)+2ml=0implies2lm=0`
`implies lm=0implies(-m-n)m=0 [ :' l=-m-n]`
`implies (m+n)m=0`
`implies m=-nimpliesm=0`
`impliesl=0,l=-n`
Thus, Dr's two lines are proportional to `0,-n,n` and `-n,0,n` i.e. `0,-1,1` and `-1,0,1`.
So the vector parallel to these given lines are `veca=-hatj+hatk` and `vecb=-hati+hatk`
Now, `cos theta=(vecavecb)/(|veca||vecb|)=1/(sqrt(2)) . 1/(sqrt(2))implies cos theta=1/2`
`:. theta =(pi)/3 [ :' "cos"(pi)/3=1/2]`