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. `(pi)/2`
B. `(pi)/3`
C. `(pi)/4`
D. `(pi)/6`

Answer

B

Explanation

We have
`l+m+n=0` and `l^(2)+m^(2)-n^(2)=0`
`impliesl^(2)+m^(2)-(-l-m)^(2)=0implieslm=0impliesl=0` or `m=0`
When`l=0`
`l+m+n=0` and `l^(2)+m^(2)-n^(2)=0`
`impliesm+n=0` and `m^(2)-n^(2)=0impliesm+n=0impliesm=-n`
So the direction ratios are proportional to `:l/0=m/1=n/(-1)`
When `m=0`
`l+m+n=0` and `l^(2)+m^(2)-n^(2)=0`
`impliesl+n=0` an `l^(2)-n^(2)=0impliesl+n=0impliesl=-n`
So the direction ratio are proportiona to `l/1=m/0=n/(-1)`
Let `theta` be the angle between the lines. Then,
`cos theta=(1xx0+1xx0+(-1)xx(-1))/(sqrt(0+1+1)sqrt(1+0+1))=1/2implies(pi)/3`