What Determines the Direction of a Normal Vector in a Plane Equation?

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SUMMARY

The direction of a normal vector in a plane equation, represented as Ax + By + Cz + D = 0, is defined by the coefficients (A, B, C). This normal vector can also be expressed in the opposite direction as (-A, -B, -C), both of which are valid. The choice of direction often depends on convenience, such as selecting a unit normal vector where |z| = 1, or on specific definitions, such as designating the outward direction on closed surfaces like spheres.

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  • Understanding of plane equations in three-dimensional space
  • Familiarity with vector mathematics
  • Knowledge of unit vectors and their significance
  • Concept of normal vectors in geometry
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For a plane with equation Ax + By + Cz + D = 0, the normal vector is (A, B, C).

However, this plane equation can also be rewritten as (-A)x + (-B)y + (-C)z + (-D) = 0, in which case the normal vector is (-A, -B, -C) which is in the opposite direction as the other normal vector.

Basically my question is this - what's the direction of a normal vector - i.e. both directions seem correct.
 
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Yes, if you simply think about the (x, y) plane, then (0, 0, 1 ) is a normal vector but so is (0, 0, -1). In fact, any vector of the form (0, 0, z) with z non-zero is a normal vector to the plane.
Which one you choose is a matter of convenience (usually |z| = 1 giving a unit normal vector is taken) and definition (if I define my z-axis to be the other way around, I can take the same normal vector but I will get a minus sign in the z-coordinate).

In some cases, for example on a sphere or other closed volume, it is common to pick the normal vector to that direction that our intuition calls "outwards", even though a vector pointing to the origin would also fit the definition.
 

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