Find the angle between the lines whose direction cosines are given by the equation `l+m+n=0` and `l^(2)+m^(2)-n^(2)=0`

Options

A. `0`
B. `pi//6`
C. `pi//4`
D. `pi//3`

Answer

D

Explanation

We have
`l+m-n=0` and `l^(2)+m^(2)-n^(2)=0`
`impliesl+m=n` an `l^(2)+m^(2)+n^(2)`
`impliesl^(2)+m^(2)=(l+m)^(2)` [On eliminating `n`]
`implies2lm=0impliesl=0` or `m=0`
If `l=0` then
`l+m-n=0impliesm=n:.l/0=m/1=n/1`
If `m=0`, then
`l+m-n=0impliesl=n:.l/1=m/0=n/1`
Thus the direction ratos of the lines are proportional to 0,1,1 and 1,0,1. Therefore angle `theta` between them is given by
`cos theta=(0xx1+1xx0+1xx1)/(sqrt(0+1+1)sqrt(1+0+1))=1/2impliestheta=(pi)/3`