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.

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?

share|improve this question
1  
the determinant of your set is 15, so it's a linear independent system. –  Max Feb 27 at 7:03
    
Such vector does not exist because they are linearly independent as Max indicates –  Semsem Feb 27 at 7:09
    
@Semsem the nullvector is orthogonal to every vector –  user127.0.0.1 Feb 27 at 7:22

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer
    
Oh, okay. So you're getting 2x + 4s + 6t = 0 by just doing row operations on the adjacency matrix? –  StarCute Feb 27 at 7:25
    
Yes. You need not use powerful matrix theory, just stick with the dot product for now. –  OC-Sansoo Feb 27 at 7:30
    
Thank you very much for your explanation. It was very helpful. –  StarCute Feb 27 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: $$ A\vec{x}=0 \implies \vec{x} = A^{-1} 0 = 0. $$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

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.