# Given two vectors, how do you get the equation of a plane that partitions the space into two parts exactly midway between the vectors?

I'm reading a book on information retrieval and in it there is an example where they have two sets of vectors.

They compute the centroid vectors for both the sets and then give the equation of a plane that divides the space into two parts equidistant from the centroid vectors between them.

I would like to know what is the method to get the equation of such a plane.

I know that given two points on a plane if you would like to find an analogous line that does the same with the points, you would connect the two points, draw a normal at the midpoint of the line connecting the points, and this normal would do the same thing that the plane does in my example. However since I have a bad background in mathematics, I can't write this mathematically, and I can't precisely think how this would work in my case.

Edit: The two centroid vectors in my example are: 0a+0b+0c+0.33d+0.33e+0.33f and 0a+0.71b+0.71c+0d+0e+0f

and the equation of the hyperplane that divides them is:

[0 -0.71 -0.71 0.33 0.33 0.33]x= -1/3

We can get the matrix on the LHS by subtracting the two given vectors, but how does this work? And I also do not know how do we get the constant on the right?

I apologize for the bad formatting, I do not know how to use TeX

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Given two points $p_1$ and $p_2$, let $v = p_2 - p_1$ be the vector from $p_1$ to $p_2$, and let $m = \frac{p_1 + p_2}{2}$ be the midpoint of $p_1$ and $p_2$. Then $$v^T(x - m) = 0$$ is the equation of a hyperplane orthogonal to $v$ that goes through $m$.
The equation I wrote could also be written $v^T x = v^T m$. Perhaps in that form it agrees with what's in your book. – littleO Oct 31 '12 at 9:40
What you call 0a+0b+0c+0.33d+0.33e+0.33f, is that just the vector $\begin{bmatrix} 0 \\ 0 \\ 0 \\ .33 \\ .33 \\ .33 \end{bmatrix}$? Also, have you given the numbers exactly as they appear in your book? – littleO Oct 31 '12 at 9:47
What's written as $.33$ is in fact $\frac13$, and $.71$ is an approximation too (though I don't see what the true value is). I suspect that if not for this approximation, we'd see perfect agreement with the formula I gave. (Though they use $v = p_1 - p_2$, an equally valid choice.) Exercise 14.15 in your book discusses this formula, which is pretty intuitive if you know how to work with and visualize vectors. – littleO Oct 31 '12 at 11:41