I Vector and Plane Relationship in 3D

Travis Enigma
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Vector contained inside the plane.
I have a quick question. If a Vector is contained inside a plane, would the normal of the plane be orthogonal to said vector?

Thank you!
 
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Yes. The normal vector of the plane is perpendicular to all vectors in the plane.
 
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Okay thank you so much!
 
fresh_42 said:
Yes. The normal vector of the plane is perpendicular to all vectors in the plane.
Does that statement relate to some axiom / definition? Pardon my non-mathematical ignorance but what constrains the normal not to be anywhere in 3D? (I'm thinking Geometry here)
 
Let ##P:=\mathbb{R}\cdot \vec{p} \oplus \mathbb{R}\cdot \vec{q}## be the plane and ##\vec{n}## its normal vector. Then ##\langle \vec{n},P \rangle =0## by definition of normality. This means that ##\langle \vec{n},\lambda \vec{p}+\mu \vec{q} \rangle=0## for all ##\lambda ,\mu \in \mathbb{R}.## This is especially true for the given vector contained in the plane whose coordinates are a specific pair ##(\lambda ,\mu).##

If you define normality by ##\vec{n} \perp \vec{p}\, \wedge \vec{n}\perp \vec{q} \,,## then we get
$$
0=\lambda \cdot 0+\mu\cdot 0=\lambda \langle \vec{n},\vec{p}\rangle +\mu \langle \vec{n},\vec{q}\rangle=\langle \vec{n},\lambda \vec{p}+\mu\vec{q}\rangle
$$
and again ##\vec{n}\perp \lambda \vec{p}+\mu\vec{q}## for a given pair of coordinates ##(\lambda ,\mu).##
 

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