The distance between the line `x=2+t,y=1+t,z=-1/2-t/2` and the plane `vecr.(hati+2hatj+6hatk)=10`, is

Options

A. `1/6`
B. `1/(sqrt(41))`
C. `1/7`
D. `9/(sqrt(41))`

Answer

D

Explanation

The equation of the lines is `(x-2)/1=(y-1)/1=(z+1/2)/(-1//2)=t`
Clearly, it is parallel to the plane `vecr.(hati+2hatj+6hatk)=10`.
So required distance is equal to the length of perpendicular form `(2,1,-1/2)` to the plane `vecr.(hati+2hatj-6hatk)-10=0` and is given by
`d=|((2hati+hatj-1/2hatk).(hati+2hatj+6hatk)-10)/(sqrt(1+4+36))|=9/(sqrt(41))`