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The equation of a plane passing through point (x₁, y₁, z₁) with normal vector (a, b, c) is
Aax + by + cz = 0
Ba(x-x₁) + b(y-y₁) + c(z-z₁) = 0
Cax₁ + by₁ + cz₁ = 0
D(x-x₁)(y-y₁)(z-z₁) = a b c
Answer & Solution
Correct answer: B. a(x-x₁) + b(y-y₁) + c(z-z₁) = 0
1. The plane's normal vector is (a, b, c).
2. Any vector from (x₁, y₁, z₁) to a general point (x, y, z) lies in the plane.
3. That vector is (x-x₁, y-y₁, z-z₁) and must be perpendicular to the normal.
4. So a(x-x₁) + b(y-y₁) + c(z-z₁) = 0.
_Source: NCERT Class 12 Maths Part 2 Ch 11 "Three Dimensional Geometry", §11.6.2_
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