The angle between the lines whose direction cosines satisfy the equations `l+m+n=""0` and `l^2=m^2+n^2` is

Options

A. `pi//6`
B. `pi//2`
C. `pi//3`
D. `pi//4`

Answer

C

Explanation

We have
`l+m+n=0`……………(i)
and `l^(2)=m^(2)+n^(2)`………………(ii)
`:.(-m-n)^(2)=m^(2)+n^(2)` [ On eliminating `l`]
`implies2mn=0impliesm=0` or `n=0`
Now,
`m=0impliesl+n=0` and `l^(2)=n^(2)` [ Putting `m=0` in `(i)` and `(ii)` ]
`impliesl=-n`
Thus, the direction ratios of one of the two lines are proportional to `-n,0,n` or `-1,0,1`
When `n=0`
`l+m+n=0` and `l^(2)+m^(2)+n^(2)`
`impliesl+m=0` an `l^(2)=m^(2)impliesl=-m`
Thus, the direction ratios of one of the two lines are poroportiona to `-m,m,0` or `-1,1,0`.
Let `theta` be the angle between the given lines. Then,
`cos theta=(-1xx(-1)+0xx1+1xx0)/(sqrt((-1)^(2)+0^(2))sqrt((-1)^(2)+1^(2)+0))=1/2impliestheta=(pi)/2`