Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on a problem I just can't seem to solve. I have three points in space, which cannot move relative to one another, and create a reference plane. There is a forth point, that lays off of the plane. How can I use the information I just described to predict where the point off the plane will lie when the plane moves? The point is constrained by the original relationship.

Thanks for any help anyone might offer!

share|improve this question
How does the plane move? How is the fourth point "constrained by the original relationship"? –  Henry Aug 1 '12 at 21:00

1 Answer 1

If the three original points (say ${\bf x}_i$ for $i=1,2,3$) are not collinear (which presumably they aren't, or else they would not determine a unique plane), then every point ${\bf z}\in\mathbb{R}^3$ can be represented uniquely as $$ {\bf z} = {\bf x}_1 + a({\bf z})({\bf x}_2 - {\bf x}_1)+b({\bf z})({\bf x}_3 - {\bf x}_1) + c({\bf z})({\bf x}_2 - {\bf x}_1)\times ({\bf x}_3- {\bf x}_1), $$ where $\times$ is the vector cross-product, and the coordinates $a,b,c\in \mathbb{R}$. These coordinates are invariant under rigid rotations and translations of $\mathbb{R}^3$; i.e., they define the position of the point relative to the three reference points only.

share|improve this answer
Should be mentioned that the point belongs to the plane $\iff c(\mathbf z)=0$. –  enzotib Aug 2 '12 at 4:15
I don't understand the a,b,c explanation , can you provide an example ? I am facing the same problem, Thanks –  ButterFly Aug 14 '12 at 6:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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