# Finding four linearly independent vectors orthogonal to each other

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?

• Are you familiar with the Gram-Schmidt process? Nov 1, 2019 at 3:18
• 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 Nov 1, 2019 at 3:21
• Yes, I just learned it a few days ago. And thanks, I'll try that.
– P123
Nov 1, 2019 at 3:28
• 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.
– amd
Nov 1, 2019 at 6:03

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.