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 3D Geometry Primer: Chapter 1 - Issue 04 - Vector Bases by (04 September 2000) Return to The Archives
 Vector Bases: A Way To Store Vectors
 We already know how to play around with vectors on 2D paper. We can draw them and make constructions. We still need to transform this to "number math" if we want to use this on a computer.

 (I) Vector Bases And Component Values

 (II) Orthogonal And Orthonormal Bases
 Until now, we used very irregular bases. There weren't special conditions, except for the linear independency of the base vectors. I'll tell you now that if you should use such a irregular base, you will have a lot of troubles incase measuring angles and distances. Orthogonal projections and decompositions of vectors will be very difficult. It is possible too, but it just ask to much efforts. Thus we need a perfect base, one that can do all that stuff without much of trouble.One step to the perfect base is when you pick all base vectors perpendicular to each other. Then you have a orthogonal base.You can do things somewhat easier already, but it ain't magic (or it *is*. It all depends on what you see as magic: the simplicity or the difficulty of math. I prefer the first one)The ideal base is when all base vectors are perpendicular to each other *and* when all base vectors have magnitude 1. Then you talk about an orthonormal base.This is what you use best in your engine if you don't want to have much math troubles. Of course, for weird effects you could use a irregular base and treat it like it's a orthonormal one. But that's for some other time. Also, there's more than one possible orthogonal or orthonormal base in your space.In short: in nD situations: irregular bases: n base vectors, linear independent. No further conditions orthogonal bases: same as irregular bases PLUS base vectors stand perpendicular to each other. orthonormal bases: same as orthogonal bases PLUS all base vectors have magnitude 1

 (III) Sequence Of Base Vectors
 There isn't a lot to tell about the sequence of base vectors, except that it does matter. What I mean is this... If you swap 2 base vectors, then you have a different base. e.g. in 3D: if you have a base B{I,J,K} and you would swap the vectors I and J, then you would have the base B'{J,I,K}.Is this important? Very! Take a vector V with component values (x,y,z) (i.e. V(x,y,z)=x*I+y*J+ z*K), and apply these values on the base B'{J, I, K}. Then you get the vector x*J+y*I+z*K. And that vector definitely differs from the original V.Especially the cross product suffers from it. The moment you need one little cross product and *bwaff*, you get in serious troubles if you don't respect the sequence of your base. For other basic operations, it doesn't really matters, but there are only a few cases in which you don't need cross products.I have to admit that I didn't realized it myself, until I learned the next thing about 3D bases: there's a difference between your lefthand and your righthand. But that's something for next week!

 (IV) Important Properties Of An Orthonormal Base
 Before we say goodnight, I want you to check out the following things about orthonormal bases. These are very important identities about base vectors, and you have to know them well. Each time you use a orthonormal base, you have to realize immediately that these things can be used. And each time you use the following stuff, you have to check for yourself if you're working in a orthonormal base.Say, we have a orthonormal base B{I,J,K}a) MAGNITUDES||I||=||J||=||K||=1Of course this is true. It's part of the definition of a orthonormal baseb) DOT PRODUCTI·J = J·I = 0 J·K = K·J = 0 K·I = I·K = 0This is true because all base vectors are perpendicular to each other (orthonormal base)I·I = J·J = K·K = 1This is also true because U·U = U² = ||U||² for each vector U, and thus also for I, J and K.c) CROSS PRODUCTI×J = K J×K = I K×I = JWhy this is will be explained next week, but you should already notice: 1. the similarity between the orthogonality of I, J, K and the cross product 2. the cyclic permutation between each rule: I becomes J, J becomes K and K becomes I again.Next week, more on this...Regards, Bramz

 Solutions
 1. A(2, 2) = 2*I + 2*J B(–2, 1) = –2*I + J C(0, –1) = –J D(–1, 2) = –I + 2*J E(–1, –2) = –I – 2*J (move E to the origin, if you don't see this) 2.

 Remarks
 (1) A set of vectors is called linear dependent if it is possible to write a least 1 vector (of that set) as a linear combination of the other vectors (of that set). A set of vectors is called linear independent if this isn't possible. (e.g. 2 vectors on the same line are linear dependent, 3 vectors on the same plane are linear dependent, 2 vectors perpendicular to each other are linear INdependent, ...)