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Find the equations of the planes parallel to the plane `x - 2y + 2z -4 = 0`, which are at a unit distance from the point (1, 2, 3).A. `x + 2y + 2z = - 6, x + 2y + 2z = 5 `B. ` x -2y - 6 = 0, x - 2y + z = 6 `C. ` x + 2y + 2z = 6, x + 2y + 2z = 0 `D. `x - 2y+ 2z = 0, x - 2y + 2z - 6 =0` |
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Answer» Correct Answer - D Given equations of plans `x - 2y + 2z + 4=0" "……..(i)` Equations 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` `rArr " "(|1-4 + 6 +k|)/(sqrt(1+4+4)) =1` `rArr " "(|3+k|)/(sqrt(9))=1` `rArr " "|3+k| = 3 rArr 3 + k = pm 3` `rArr" "k = 0,-6` `thereofre ` Required equation of plane are `x - 2y + 2 +0=0 and x -2y + 2z - 6 =0 ` `rArr x - 2y + 2z = 0 and x -2y + 2z - 6 =0` |
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