# 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!

Mentor
2022 Award
Yes. The normal vector of the plane is perpendicular to all vectors in the plane.

• jedishrfu and Travis Enigma
Travis Enigma
Okay thank you so much!

Gold Member
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)

Mentor
2022 Award
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).##