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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. . . |
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Answer» x+2y+2z=-6,x+2y+2z=5 `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` |
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