Vector and Plane Relationship in 3D

[Travis Enigma]

TL;DR Summary

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!

 

[fresh_42]

Yes. The normal vector of the plane is perpendicular to all vectors in the plane.

 

[Travis Enigma]

Okay thank you so much!

 

[sophiecentaur]

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)

 

[fresh_42]

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 $(\lambda ,\mu).$