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How to write an equation for plane, which includes dots with radius-vectors $\mathbf r_{1}, \mathbf r_{2}, \mathbf r_{3}$ that do not lie on a straight line? The answer is

$ (\mathbf r, ([\mathbf r_{3} , \mathbf r_{1}] + [\mathbf r_{2} , \mathbf r_{3}] + [\mathbf r_{1} , \mathbf r_{2}]) ) = (\mathbf r_{1}, [\mathbf r_{2}, \mathbf r_{3}]). $

How to explain it?

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    Maybe, there was used a complanarity of vectors, which are a combination of $\mathbf r_{1}, r_{2}, r_{3}$. So, that's like $((\mathbf r - \mathbf r_{1}), [(\mathbf r_{2} - \mathbf r_{1}), (\mathbf r_{3} - \mathbf r_{1})]) = 0,$ if I'm right.2012-12-04

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Thanks to the comments, I think I understand the problem: show that the given formula is an equation for the plane containing the three non-collinear points $r_1,r_2,r_3$. Given a general point $r$ in the plane, $r-r_1$ will be orthogonal to the cross product of $r_2-r_1$ and $r_3-r_1$. That is, $(r-r_1)\cdot((r_2-r_1)\times(r_3-r_1))=0\tag1$ which is to say $r\cdot((r_2-r_1)\times(r_3-r_1))=r_1\cdot((r_2-r_1)\times(r_3-r_1))\tag2$ Now $v\times v=0$ for all $v$, so $(r_2-r_1)\times(r_3-r_1)=r_2\times r_3-r_2\times r_1-r_1\times r_3\tag3$ Also, $u\times v=-v\times u$, and $u\cdot(u\times v)=0$, so (2) becomes $r\cdot(r_2\times r_3+r_1\times r_2+r_3\times r_1)=r_1\cdot(r_2\times r_3)\tag4$ and we're done.