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?

  • 9
    $\begingroup$ Two (independent) vectors determine a unique plane through the origin. $\endgroup$ Oct 6 '14 at 4:07
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 6 '14 at 4:22
  • 1
    $\begingroup$ 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. $\endgroup$
    – user541686
    Oct 6 '14 at 9:39
  • $\begingroup$ @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 ;-) $\endgroup$ Oct 6 '14 at 15:12

Remember, vectors don't have starting or ending positions, just directions. So take the vectors <1,0,0> and <0,1,0>. These vectors will define a plane that only goes in the x-y direction, but the problem is, they will work for any z-coordinate. So you need some starting point to anchor your plane.

  • $\begingroup$ Ah...yes of course, it's very clear now, thank you! $\endgroup$ Oct 6 '14 at 4:10
  • 6
    $\begingroup$ no problem, good that you're thinking things over and not just taking things at face value. $\endgroup$
    – Alan
    Oct 6 '14 at 4:12
  • 1
    $\begingroup$ I don't know how it is in general math, but as a programmer I wouldn't know what to do with just two vectors, while a point and a normal are the basis to pretty much all plane related code. $\endgroup$
    – Kevin
    Oct 6 '14 at 14:56
  • $\begingroup$ @Kevin: two non-colinear vectors are literally a so-called "basis" for a unique plane through the origin. You can express any point in the plane as a linear combination of the two vectors, and/or you can take their cross product (and make it unit length if necessary) to get a normal to the plane and do whatever you'd have done with a normal. But this of course relies on first choosing the origin or some other point in the plane to establish which it is of many parallel planes. $\endgroup$ Oct 6 '14 at 15:06
  • 1
    $\begingroup$ Just wanted to thank people for my first nice answer badge. I feel all appreciated! (Now to go torture the methods of calc students who are my actual students this semester.....) $\endgroup$
    – Alan
    Oct 6 '14 at 15:29

There are infinitely many planes parallel to two linearly independent vectors.

  • $\begingroup$ but isn't that also true of the xy-plane? $\endgroup$ Oct 6 '14 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.