Sunday, January 28, 2024

GeoTheorem+ 1: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on

Geometry+ GeoTheorem 1


Prove that non-null vector n(a,b,c) in three-dimensional Cartesian coordinates OXYZ is normal (perpendicular) to a plane α described by an equation
a·x + b·y + c·z + d = 0
where a, b, c, d are real numbers.


CASE 1 (easy) - Constant d in an equation that describes plane α equals to zero.

The equation for plane α looks in this case as
a·x + b·y + c·z = 0

Then plane α must go through the origin of coordinates O(0,0,0) because this point satisfies the equation for α.

Consider a vector from an origin of coordinate O(0,0,0) to any other point on a plane Q(x,y,z).
Obviously, non-null vector OQ(x,y,z) is lying fully within plane α because both its ends - points O and Q lie within this plane.

We can interpret the equation
a·x + b·y + c·z = 0
that describes plane α in this case as a scalar product of non-null vector n(a,b,c) and non-null vector OQ(x,y,z).

Since this scalar product between vector n(a,b,c) and any vector OQ(x,y,z) lying within plane α is zero, vector n(a,b,c) is perpendicular to plane α.

CASE 2 - Constant d is not equal to zero.

Consider two planes defined by two equations
Plane α: a·x + b·y + c·z + d = 0
Plane β: a·x + b·y + c·z = 0

Since d ≠ 0, any point that satisfies one of these equations will not satisfy another.
Therefore, these planes do not have any common points and, therefore, are parallel.

We have already proven that non-null vector n(a,b,c) is perpendicular to plane β (see CASE 1 above).
Consequently, this vector n is perpendicular to α as well, that is n is normal to α.

End of the proof that nα.

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