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Question
Find the perpendicular distance of origin from the plane 6x − 2y + 3z - 7 = 0
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Solution
Given equation of plane is 6x – 2y + 3z – 7 = 0
Comparing the given equation with ax + by + cz + d = 0,
we get a = 6, b = – 2, c = 3, d = – 7
∴ Direction cosines are,
l = `6/sqrt(6^2 + (-2)^2 + 3^2)`
m = `(-2)/sqrt(6^2 + (-2)^2 + 3^2)`
n = `3/sqrt(6^2 + (-2)^2 + 3^2)`
∴ l = `6/7, "m" = (-2)/7, "n" = 3/7`
Cartesian equation of a plane in normal form is lx + my + nz = p
∴ `6/7x - 2/7y + 3/7z = 7/7`
∴ `6/7x - 2/7y + 3/zz` = 1
∴ The distance of the origin from the plane is 1.
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