Let `P=-(1,7,sqrt(2))` be a point and line L is `2sqrt(2)(x-1)=y-2,z=0`. If PQ is the distance of plane `sqrt(2)x+y-z=1` from point P measured along a line inclined at an angle of `45^(@)` with the line L and is minimum then the value of PQ is

Options

A. 3
B. 4
C. 6
D. 8

Answer

A

Explanation

Line L is `(x-1)/1=(y-2)/2sqrt(2)=z/0`
This line L makes an angle of `45^(@)` with the plane `sqrt(2)x+y-z=1`
`therefore` Required distance PQ is prependicular distance of plane from P
I.e., `PQ=(|sqrt(2)+7-sqrt(2)-1|)/sqrt(2+1+1)=3`