# Proof about subspaces in $R^3$

Problem : Prove that each plane that intersects the origin of the vector space $R^3$ is a subspace of $R^3$

I've tried to tackle this by showing that the subspace is not empty, that is closed under vector addition and scalar multiplication, but I'm not sure how to define the subspace mathematically.

• What does "plane that moves through the origin" mean? – Spenser Feb 25 '16 at 20:57
• What is the equation of a plane? What is the equation of a plane that intersects the origin? – Peter Franek Feb 25 '16 at 20:57
• From what I can remember, you're done. A "subspace" is a "(proper) subset of a vector space who also happens to be a vector space itself" so, as long as your demonstration does not assume that the vectors are out of $R^3$ there's nothing left to demonstrate. – polettix Feb 25 '16 at 21:01

a plane in $R^3$ has a unique(except for a sign) unit normal, say, $(a,b,c).$ the equation of the plane is $\{(x,y,z): ax+by+cz = 0\}$ show that if two points in the plane then their sum and scalar multiples are also on the plane.
A subvector space of a vector space $V$ over an arbitrary field $\mathbb{F}$ is a subset $U$ of $V$ which contains the zero vector and for any $v,w \in U$ and any $a,b \in \mathbb{F}$ it is the case that $av+bw \in U$, so the equation of the plane in $\mathbb{R^3}$ parallel to $v$ and $w$, and containing the origin is of the form $$x=a v_{1} +b w_{1}$$ $$y=a v_{2} +b w_{2}$$ $$z=a v_{3} +b w_{3}$$ where $v=(v_1,v_2, v_3)$ $w=(w_1, w_2, w_3)$ which means that all the points (which can be identified with vectors) in that plane are linear combinations of $v$ and $w$ (i.e. are of the form $av+bw$) and you can also verify that the origin is in that plane