3D Geometry Primer: Chapter 1 - Issue 10 - Addendum: What I Didn't Tell You (Yet) by (16 October 2000) Return to The Archives
 Addendum: What I Didn't Tell You (Yet) Right, this really is the last issue of the 1st chapter. I've been telling you this already for a few issues, but now it's really the end. But don't be mournful, in a few weeks/months, there will be chapter II :)Today, I'll tell you things I forgot to tell you in the past issues. I'll give you answers to questions that you asked me, and give you answers I've got on questions I've asked. But, let's start with a question of yours, which I will turn into a new question :)

 (I) Why The Dot And Cross Products Are What They Are ... (II) Areas Of Triangles And Volumes Of Tetrahedras a) Areas Of TrianglesHow do you compute the area of a triangle? Well, I hope you remember the following picture: the area of the triangle ABC is exactly the half of the parallellogram ABDC: a = (b*h)/2 But then, of course, you must know the base b and height h of the triangle, which isn't always that easy if you just have the coordinates of the vertices A, B and C.But here comes the trick: if you take the magnitude of the cross product (AB)×(AC), then you become exactly the area of the parallelogram enclosed between the vectors AB and AC. So, if you take the half of it, you become the area of you triangle ABC. Area a of triangle ABC = (1/2)*||(B–A)×(C–A)||(Remark: the notation AB is just a way to talk about the free vector that fits between points A and B, which can be calculated by: AB = B – A)b) Volumes of tetrahedras I don't know if you know this, but if you want to calculate the volume v of a pyramid ABCD (which the tetrahedra is), then you take 1/3 of the volume of a prism ABC DEF with the same base surface and height. It's possible to prove this, but that's something with integrals. So, I won't do that here. But that prism is also exactly the half of the parallelepiped ABCG DEFH. And the volume of that parallelepiped can be calculated with the triple product (AB AC AD). In that way, it's possible to compute the volume of the tetrahedra ABCD by taking 1/6 of the volume of the parallelepiped. Don't forget to take the absolute value of the triple product, since it can be positive or negative. Volume v of tetrahedra ABCD = abs( (1/6)*(AB AC AD) ) = abs( (1/6) * ((B–A)×(C–A))·(D–A) )

 (III) Why Is The Origin Of The Screen In The Upper-Left Corner? (IV) Free Vectors As Linear Combinations Of Points In the same way as it's possible to set conditions to weight-factors w[i] in a lin. comb. of points to achieve a new point (independent to the origin); it's possible to become a free vector as a lin. comb. of points (also independent to the origin). How? First, let's write down what we want: You can use an analogue reasoning as in the last issue to become the condition on w[i], but I'll leave that as an exercise to the reader. Here, I will tell you a more intuitive reasoning:At the right side, you see w[i]*P[i] which you will write as w[i]*OP[i]. This place vector contains the origin O and the point P[i]. This means that for each point P[i] at the right side, you will also have a w[i]*O. If you summate this, you will have a Sw[i]*O at the right side.But at the left side, you only have the already free vector V. No origin O is in there. So, if you want both sides to be equal, then you may not have an origin in the right side too. This is why Sw[i] must be 0 (zero).In one line, this becomes: (V) Splitting Up Barycentric Combinations It's always possible to split up a barycentric comb. in a convex barycentric comb. (in which all w[i]ł0) and a "free vector comb." (in which Sw[i]=0). This is very easy to prove if you take the barycentrum (w[i]=1/n) as convex barycentric comb. Then each barycentric combination can be written as the sum of the barycentrum + a free vector: Convince yourself that this is correct. The orange part is the barycentrum (which is convex) and the yellow part is the free vector.In some situations, it's possible to split this combinations in a better way. I'll give you an example. Say, you have the combination Q = (0.5)*A + B – C + (0.5)*D. You can write this as: Q = ((0.5)*A + (0.5)*D) + (B – C), which is the middle of the points A and B *plus* the free vector CB. Now you immediately see what Q will be.I don't know where you can use this, but it shows you that each barycentric combination can be written as the sum of a convex barycentric comb. + a free vector.

 (VI) Clifford Algebra Many many weeks ago, Jaap Suter passed me a link to an algebra site with an introduction on Clifford Algebra. I said that I would put this link in my primer one day. I think that, no, I'm sure that, the stuff explained there, will be very scary for most of you, but hey I've promised Jaap :) So here it is: Introduction to Geometric Algebra. In fact, I'm already a few weeks too late, since Tim already posted this in the news section.This is it. The last word of chapter one ... There's nothing more to say than a 'goodbye', so: "goodbye and CU in chapter II".Regards, Bramz"And what I see, you can not see. That when I move, you move with me" - Ribeiro (Moonspell)