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Question
If the normal to the plane has direction ratios 2, −1, 2 and it’s perpendicular distance from origin is 6, find its equation
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Solution
Direction ratios of the normal to the plane are 2, −1, 2
∴ a = 2, b = –1 and c = 2
∴ Direction cosines are,
l = `2/sqrt(2^2 + (-1)^2 + 2^2)`
m = `(-1)/sqrt(2^2 + (-1)^2 + 2^2)`
n = `2/sqrt(2^2 + (-1)^2 + 2^2)`
∴ l = `2/3, "m" = (-1)/3, "n" = 2/3`
Cartesian equation of a plane which is at the distance of 6 units from the origin is
lx + my + nz = 6
∴ `2/3x - 1/3y + 2/3z` = 6
∴ 2x – y + 2z = 18
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