# Determining all the vectors which are orthogonal to given vector in $\Re^3$

Determine all the vectors [x y z] which are orthogonal to [1 2 3], [-2, 1, -1] and [0 -1 2]. My work so far: $\bigl(\begin{smallmatrix} x\\ y\\ z \end{smallmatrix} \bigr)$ $\bigl(\begin{smallmatrix} 1\\ 2\\ 3 \end{smallmatrix} \bigr)$ = 1x + 2y + 3z = 0 $\to$ 3z = 1x + 2y. Let s = x and t = y $\to$ 3z = 1s + 2t. $\bigl(\begin{smallmatrix} 1\\ 2\\ 3 \end{smallmatrix} \bigr)$ $\bigl(\begin{smallmatrix} s\\ t\\ s+2t \end{smallmatrix} \bigr)$ If this is right I'm not sure where to go from here. I know that two vectors are orthogonal if their dot products are equal to zero, but how can I know "all" the vectors that are orthogonal to a given vector given what I have done so far?

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the determinant of your set is 15, so it's a linear independent system. – Max Feb 27 '14 at 7:03
Such vector does not exist because they are linearly independent as Max indicates – Sameh Shenawy Feb 27 '14 at 7:09
@Semsem the nullvector is orthogonal to every vector – user127.0.0.1 Feb 27 '14 at 7:22

let u = (1,2,3), v = (-2,1,-1), and w = (0,-1,2) and x = (r, s, t), then : u*x = 0 ==> r + 2s + 3t = 0, v*x = 0 ==> -2r + s - t = 0, and w*x = 0 ==> -s + 2t = 0. The 1st equation gives: 2r + 4s + 6t = 0 . Adding this with the 2nd equation gives: 5s + 5t = 0 so s + t = 0 , and together with the 3rd equation gives: 3t = 0 ==> t = 0, and s = 0, and r = 0. So the only "vector" that is orthogonal to all three given vectors is the zero vector.

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Oh, okay. So you're getting 2x + 4s + 6t = 0 by just doing row operations on the adjacency matrix? – StarCute Feb 27 '14 at 7:25
Yes. You need not use powerful matrix theory, just stick with the dot product for now. – DeepSea Feb 27 '14 at 7:30
Thank you very much for your explanation. It was very helpful. – StarCute Feb 27 '14 at 7:48

Think about what you're being asked -- you have 3 dot products you want to be 0. We can write this explicitly as a matrix operation: $$\begin{pmatrix} 1&2&3\\-2&1&-1\\0&-1&2\end{pmatrix} \vec{x} = 0$$

We want to find $\vec{x}$. As was remarked by @Max, the determinant of this matrix is 15; more importantly, it is not 0. Therefore, this matrix can be inverted and we have a unique solution. Of course, this solution must be 0, because: $$A\vec{x}=0 \implies \vec{x} = A^{-1} 0 = 0.$$

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To say it in still another way, given that the three vectors you start with are linearly-independent, if there was some (EDIT: non-zero) vector $v$ that was orthogonal to these three, then you would have four linearly-independent vectors in $\mathbb R^3$ , a 3-dimensional vector space. So only $(0,0,0)$ can be oerthogonal to all.

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