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I came up with these vectors:

u = \begin{bmatrix}1\\ 1\\ 0 \\1 \\ \end{bmatrix} v = \begin{bmatrix}1\\ -1\\ 1 \\0 \\ \end{bmatrix} w = \begin{bmatrix}2\\ -2\\ -4 \\0 \\ \end{bmatrix}

u and v are orthogonal. w is orthogonal to u and v. I have to find another matrix z that's orthogonal to u, v, and w, and I'm having a lot of trouble. Is there a better or more efficient way to do this than to simply keep guessing random vectors and adjusting them?

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    $\begingroup$ Are you familiar with the Gram-Schmidt process? $\endgroup$
    – Math1000
    Nov 1, 2019 at 3:18
  • $\begingroup$ Welcome to Mathematics Stack Exchange. If you can find a fourth vector linearly independent of the three, then you can use the Gram-Schmidt process to find one that's orthogonal to the three $\endgroup$
    – J. W. Tanner
    Nov 1, 2019 at 3:21
  • $\begingroup$ Yes, I just learned it a few days ago. And thanks, I'll try that. $\endgroup$
    – P123
    Nov 1, 2019 at 3:28
  • $\begingroup$ No need for G-S here. The dot products of $z$ with each of the other three vectors must vanish. This gives you a system of homogeneous linear equations, the solution to which is the null space of a certain $3\times4$ matrix. $\endgroup$
    – amd
    Nov 1, 2019 at 6:03

1 Answer 1

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In four dimensional euclidean space, the standard unit vectors $(1,0,0,0)\,,(0,1,0,0)\,,(0,0,1,0)$ and $(0,0,0,1)$ satisfy your condition.

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