2

I want to check if three vectors are in the same plane, the vectors being $$A(2,-1,2),B(1,2,-3),C(3,-4,7). $$

What I did so far is to create vector $AB ( -1,3,-5)$ and build the plane equation with the point $A$ $$-1(x-2)+3(y+1)-5(z-2)=0$$ and inserted the point $C$ to check if the equation exists. Is this the right way to do that or I did something wrong?

Thanks!

Ofir Attia
  • 3,136

2 Answers2

6

It would be easier to compute a scalar triple product.. Make three vectors $\hat{u}, \hat{v},\hat{w}$.
The three vectors lay in the same plane iff $$ \hat{u} \cdot (\hat{v} \times\hat{w} ) = 0$$

  • 1
    This doesn't seem to be correct: three different points in space do always lie in a plane...either the OP meant a plane through the origin or something else. – DonAntonio Jul 25 '13 at 07:32
  • 2
    @DonAntonio: the OP speaks of vectors ... – Tony Piccolo Jul 25 '13 at 08:12
  • I see @TonyPiccolo, but then are we to assume they all are "anchored" at the same point, say the origin? And when dealing with advanced analytic geometry the distinction between "points" and "vectors" is not that clear, in particular after a little linear algebra kicks in...Anyway, the above answer says "make three vectors..." What does that mean in this context: if we already have vectors, what vectors are there to be made? And if we do not have "vectors", then how are are those to be made? – DonAntonio Jul 25 '13 at 08:16
  • 1
    @DonAntonio: one has to check whether $(\alpha,\beta,\gamma)\neq(0,0,0)$ exists such that $\alpha \hat{u}+ \beta \hat {v}+ \gamma \hat{w}=0$, i.e. the vectors are linearly dependent. – Tony Piccolo Jul 25 '13 at 08:28
  • I'm not sure I follow you, Tony...but how's that related to this asnwer being, imfho, incorrect? – DonAntonio Jul 25 '13 at 08:30
  • 1
    @DonAntonio: In this search coplanar. – Tony Piccolo Jul 25 '13 at 08:40
  • Thanks @TonyPiccolo, I did know that though. My question (all along) was how all this makes this answer correct? What are those misterious (to me) vectors $,\hat u...$ in the answer? If it was meant to be the given ones, why give them new names and choose the word "make"? I think this, together with the question itself, isn't that clear. – DonAntonio Jul 25 '13 at 08:44
  • @DonAntonio: I agree, it should have been better to say "given" instead of "make". However John Doe gives a statement that doesn't need to be related to the vectors in the OP's question. – Tony Piccolo Jul 25 '13 at 08:56
  • 2
    There are a few "not an answer" flags on this. It looks like people might be flagging it because they think some part is technically wrong. That flag is not appropriate in such a case: a few reasonably good applications of the "not an answer" flag include when the OP posts new questions as answers, when people make irrelevant comments as answers, or when an answer is just gibberish. – rschwieb Jul 25 '13 at 12:42
  • @rschwieb I strongly concur with this. – Shuhao Cao Jul 25 '13 at 15:00
  • @rschwieb I officially concur with this. – Alex Becker Jul 25 '13 at 21:50
0

If $\alpha, \beta,\gamma\in \mathbb{R}^3,$ then we have: $\alpha,\beta,\gamma$ are linearly independent $\iff$ the determinant of matrix $(\alpha,\beta,\gamma)$ is not zero, where $\alpha,\beta,\gamma$ are regarded as column vectors. And we also know that that $\alpha,\beta,\gamma $ is linearly dependent is equivalent to that $\alpha,\beta,\gamma$ are in the same plane. Then it is easy to check $$\begin{vmatrix}2 &-1&2\\1 &2 &-3\\ 3 &-4 &7 \end{vmatrix}=0,$$ so you can draw a conclusion that they are coplanar.

nuage
  • 916