1.

The shortest distance between the lines `2x + y + z - 1 = 0 = 3x + y + 2z - 2` and `x = y = z`, isA. `1/(sqrt(2))`B. `sqrt(2)`C. `3/(sqrt(2))`D. `(sqrt(3))/2`

Answer» Correct Answer - A
The shortest distance between the given lines is equal to the length of the perpendicular from any point on `(x-0)/y=(y-0)/1=(z-0)/1` to the plane through the line `2x+y=z-1=10, 3x+y+2z-2=0` and parallel to `(x-0)/1=(y-0)/1=(z-0)/1`
The equation of any plane through the line `2x+y+z-1=10, 3x+y+2z-2=0` is
`(2x+y+z-1)+lamda(3x+y+2z-2)=0`
or `x(3lamda+2)+y(lamda+1)+z(2lamda+1)-2lamda-1=0`...........i
If it is parallel to `(x-0)/1=(y-0)/1=(z-0)/1`, then
`3lamda+2+lamda+1+2lamda+1=0implieslamda=(-2)/3`
Putting `lamda=(-2)/3` in (i) we obtain `-y+z-1=0`
The distance of this plane from any point on `(x-0)/1=(y-0)/1=(z-0)/1` is
`d=|(-0+0-1)/(sqrt(1+1))|=1/(sqrt(2))`


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