The `vec b` which is collinear with the vector `vec a = (2,1,-1)` and satisfies the relation `vec a. vecb =3` is

Options

A. (1, 1/2, -1/2)
B. (2/3, 1/3, -1/3)
C. (1/2, 1/4, -1/4)
D. (1,1,0)

Answer

A

Explanation

Let `vec(b)=x vec(i)+y vec(j)+z vec(k)`
Since, `vec(b)` is collinear with vector `vec(a)`
therefore `vec(a)=k vec(b)` where k is a scalar.
Given `vec(a)=(2, 1, -1)`
`:. (2, 1, -1)=k(x, y, z)`
`rArr x=(2)/(k), y=(1)/(k), z=(-1)/(k)`
Also, `vec(a). vec(b)=3`
`rArr 2x+y-z=3`
`2((2)/(k))+(1)/(k)+(1)/(k)=3`
`rArr (4)/(k)+(1)/(k)+(1)/(k)=3`
`rArr (6)/(k)=3 rArr k=2`
`:. x=1, y=(1)/(2) and z=(-1)/(2)`
Hence `vec(b)=(1, (1)/(2), (-1)/(2))`