1.

Equations of planes parallel to the plane x-2y+2z+4=0 which are at a distance of one unit from the point (1,2,3) are. . .

Answer»

x+2y+2z=-6,x+2y+2z=5
x-2y-6=0,x-2y+z=6
x+2y+2z=6,x+2y+2z=0
x-2y+2z=0,x-2y+2z-6=0

Solution :Given equation of plane
`x-2y+2z+4=0`. . (i)
Equation of plane parallel to plane (i) is
`x-2y+2z+K=0`. . .(ii)
since, DISTANCE of plane (ii) from point (1,2,3) is 1 unit
`therefore(|1(1)-2(2)+2(3)+k|)/(sqrt((1)^(2)+(-2)^(2)+(2)^(2)))=1`
`IMPLIES(|1-4+6+k|)/(sqrt(1+4+4))=1 `
`implies(|3+k|)/(sqrt(9))=1 `
`implies|3+k|=3implies3+k=+-3impliesk=0,-6`
`therefore`Required equation of plane are
`x-2y+2z+0=0 and x-2y+2z-6=0`
`impliesx-2y+2z=0 and x-2y+2z-6=0`


Discussion

No Comment Found

Related InterviewSolutions