# Why not use two vectors to define a plane instead of a point and a normal vector?

In Multivariable calculus, it seems planes are defined by a point and a vector normal to the plane. That makes sense, but whereas a single vector clearly isn't enough to define a plane, aren't two non-colinear vectors enough?

What I'm thinking is that since we need the cross-product of two vectors to find our normal vector in the first place, why not just use those two vectors to define our plane. After all, don't two non-colinear vectors define a basis in R2?

• Two (independent) vectors determine a unique plane through the origin. – André Nicolas Oct 6 '14 at 4:07
• Two linearly independent vectors do give rise to the parametrization of a plane containing point $p$. Simply use $\vec{r}(u,v)=p+u\vec{A}+v\vec{B}$. Sometimes, this is more useful than the Cartesian equation of a plane, it's all about context. – James S. Cook Oct 6 '14 at 4:22
• You really only need 1 vector and 1 real number at most... I'm trying to see if I can reduce that down to 1 vector, and in most c cases that's possible too, so 2 vectors itself is clearly redundant. – user541686 Oct 6 '14 at 9:39
• @Mehrdad: you can almost just use the length of the vector in place of the real number. The problem is that the direction vanishes when the length is 0, so you get one point (presumably the origin), such that the planes through that point can't be distinguished. Ultimately though it's a matter of neatness. If you let your problem become "is there a bijection between 3-vectors and 2-planes in 3-space" then yes, of course, they both have the cardinality of the reals ;-) – Steve Jessop Oct 6 '14 at 15:12