Height and DistanceHard
Question
A plane passes through (1, - 2, 1) and is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance of the plane from the point (1, 2, 2) is
Options
A.0
B.1
C.√2
D.2√2
Solution
The plane is a(x - 1) + b(y + 2) + c(z - 1) = 0
where 2a - 2b + c = 0 and a - b + 2c = 0
⇒
So, the equation of plane is x + y + 1 = 0
∴ Distance of the plane from the point (1, 2, 2) =
= 2√2.
where 2a - 2b + c = 0 and a - b + 2c = 0
⇒
So, the equation of plane is x + y + 1 = 0
∴ Distance of the plane from the point (1, 2, 2) =
= 2√2.Create a free account to view solution
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