Finding Orthogonal Vectors in 4 Space

[shane1]

I have this question that says:
Find two vectors of norm 1 that are orthagonal to the three vectors u = (2, 1, -4, 0), v = (-1, -1, 2, 2), and w = (3, 2, 5, 4).

I've tried setting up a system of equations to solve.
2a + b - 4c = 0
-a - b + 2c + 2d = 0
3a + 2b + 4c + 4d = 0

But when I did that I was left with a free variable. So basically I was wondering if there's another way to do it such as taking the determinate like how you do in 3 space. Except in 4 space.
Eg.
i j k
0 1 0
1 2 5

Shane

 

[StatusX]

There will be a whole line of vectors perpendicular to those vectors. But only 2 will have norm 1.

 

[NateTG]

If you know how to calculate the determinat of an nxn matrix there is an n-dimensional analog of the cross product:
<br /> \vec{v}=\left| \begin{array}{c c c c}<br /> \hat{i} &amp; \hat{j} &amp; \hat{k} &amp; \hat{l} \\<br /> 2 &amp; 1 &amp; -4 &amp; 0 \\<br /> -1 &amp; -1 &amp; 2 &amp; 2 \\<br /> 3 &amp; 2 &amp; 5 &amp; 4 \end{array} \right |

Which will give you a vector perpendicular to the n-1 you already have.