[Math] Necessary and sufficient conditions for a point to lie inside a tetrahedron (vectors)

geometryvectors

I am given the problem

Find a necessary and sufficient condition for a point to lie inside or on the tetrahedron formed by the position vectors O ,a, b, c.

In an earlier part of the question I was asked to find an equation of the plane passing through a,b and O which is r . (axb)=0, and then state a condition that, for a point above or below the plane, the point will be on the same side of the plane as a vector c. For this part I considered that the normal to plae on a particular side is axb, so for r and c on the same side it must be that r.(axb) and c.(axb) are the same sign. I am also given that a.(bxc)>0, so a condition must be that r.(axb)>0

In the next part of the question I am asked to show that the equation of a plane through a,b and c is

r.(axb + bxc + cxa)=a.(bxc)

which was simple enough to do.

I have a feeling that these parts are needed for the question I asked at the beginning, which is why I put them here.

Now my first thought was that if I gave a condition for a point being on or above each of the faces, then it must be contained within the tetrahedron. Using this approach, for the faces passing through the origin I get

1. r.(axb)$\geq$0

2. r.(bxc)$\geq$0

3. r.(cxa)$\geq$0

Now for the final face which does not pass through the orgin I am not sure, and even if my approach is correct I am pretty sure it is not the most elegant way to think about this; I considered the position vector a as a new 'origin'. I know that the normal to this plane is (axb + bxc + cxa) and I need a point which lies on the same side of this plane as the actual origin, which is a position vector -a from a

So using the same approach as before I need

(r-a).(axb + bxc + cxa) to be the same sign as (-a).(axb + bxc + cxa)=-a.(bxc). So I need (r-a).(axb + bxc + cxa) to be negative, meaning

4. r.(axb + bxc + cxa)$\leq$a.(bxc)

So I am not sure if this is at all correct, and i'm pretty sure there is a better way to think about this. I have come across something like this before and I think the linear independence of the vectors was used? Perhaps I could say that r= pa+qb+sc where p+q+s=1? I am not sure about this though. In some ways it seems to make sense, but it is not at all intuitive. And also somehow this has only one (or two?) conditions, whereas I had four before! So I don't really understand if either of these answers is correct. Unfortunately I don't seem to have a very good intuition when it comes to geometry.

Best Answer

Your last hypothesis on a linear combination of the given vectors is .. on the right track. Since you are asking to provide an intuitive explanation let's go by steps.
Consider the two vectors $\mathbf a$ and $\mathbf b$ as position vectors with respect to $O$: so they individuate two points $A$ and $B$ in the given reference system.
Now

  • you should akready know that $\mathbf {r} = \lambda \mathbf {a} +(1-\lambda) \mathbf {b}$ ($\mathbf r$ being the vector $(x,y,z)$ as per your notation and $\lambda$ a scalar) represents a line through $A,B$, and that when $0 \leq \lambda \leq 1$ you get the segment $[A,B]$. That is the same as $$ \mathbf{r} = \lambda \mathbf{a} + \mu \mathbf{b}\quad \left| {\;\left\{ \begin{gathered} 0 \leqslant \lambda ,\mu \left( { \leqslant 1} \right) \hfill \\ \lambda + \mu = 1 \hfill \\ \end{gathered} \right.} \right. $$
  • if you replace $\lambda + \mu = 1$ with $\lambda + \mu \leqslant 1$ then you clearly get the points inside (and on the border of) the triangle $\triangle OAB$ ;
  • once the above is clear, you can get that $$ \mathbf{r} = \lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}\quad \left| {\;\left\{ \begin{gathered} 0 \leqslant \lambda ,\mu ,\nu \left( { \leqslant 1} \right) \hfill \\ \lambda + \mu + \nu = 1 \hfill \\ \end{gathered} \right.} \right. $$ represents the triangle $\triangle ABC$ ($\mathbf{r} - \mathbf{a} = \left( {\lambda + \mu + \nu - 1} \right)\mathbf{a} + \mu \left( {\mathbf{b} - \mathbf{a}} \right) + \nu \left( {\mathbf{c} - \mathbf{a}} \right) $), and that putting instead $\lambda + \mu + \nu \leq 1$ is the result you are looking for.
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