Htykuut

Suppose we have a vector $(a,b)$ in $2$-space. Then the vector $(-b,a)$ is orthogonal to the one we started with. Furthermore, the function $$(a,b) mapsto (-b,a)$$ is linear.

Suppose instead we have two vectors $x$ and $y$ in $3$-space. Then the cross product gives us a new vector $x times y$ that's orthogonal to the first two. Furthermore, cross products are bilinear.

Question. Can we do this in higher dimensions? For example, is there a way of turning three vectors in $4$-space into a fourth vector, orthogonal to the others, in a trilinear way?

Yes. It's just like in dimension $3$: if your vectors are $(t_1,t_2,t_3,t_4)$, $(u_1,u_2,u_3,u_4)$, and $(v_1,v_2,v_3,v_4)$, compute the formal determinant:$$begin{vmatrix}t_1&t_2&t_3&t_4\u_1&u_2&u_3&u_4\v_1&v_2&v_3&v_4\e_1&e_2&e_3&e_4end{vmatrix}.$$ You then see $(e_1,e_2,e_3,e_4)$ as the canonical basis of $mathbb{R}^4$. Then the previous determinant is $(alpha_1,alpha_2,alpha_3,alpha_4)$ withbegin{align*}alpha_1&=t_4u_3v_2-t_3u_4v_2-t_4u_2v_3+t_2u_4v_3+t_3u_2v_4-t_2u_3v_4\alpha_2&=-t_4u_3v_1+t_3u_4v_1+t_4u_1v_3-t_1u_4v_3-t_3u_1v_4+t_1u_3v_4\alpha_3&=t_4u_2v_1-t_2u_4v_1-t_4u_1v_2+t_1u_4v_2+t_2u_1v_4-t_1u_2v_4\alpha_4&=-t_3u_2v_1+t_2u_3v_1+t_3u_1v_2-t_1u_3v_2-t_2u_1v_3+t_1u_2v_3end{align*}It's a vector orthogonal to the other three.

I followed a suggestion taken from the comments on this answer: to put the entries $e_1$, $e_2$, $e_3$, and $e_4$ at the bottom. It makes no difference in even dimension, but it produces the natural sign in even dimension.

Following another suggestion, I would like to add this remark:$$alpha_1=-begin{vmatrix}t_2&t_3&t_4\u_2&u_3&u_4\v_2&v_3&v_4end{vmatrix}text{, }alpha_2=begin{vmatrix}t_1&t_3&t_4\u_1&u_3&u_4\v_1&v_3&v_4end{vmatrix}text{, }alpha_3=-begin{vmatrix}t_1&t_2&t_4\u_1&u_2&u_4\v_1&v_2&v_4end{vmatrix}text{ and }alpha_4=begin{vmatrix}t_1&t_2&t_3\u_1&u_2&u_3\v_1&v_2&v_3\end{vmatrix}.$$

My answer is in addition to José's and Antinous's answers but maybe somewhat more abstract. In principal their answers are using coordinates whereas I'm trying to do it coordinate-free.

What you are looking for is the wedge or exterior product.
The exterior power $bigwedge^k(V)$ of some vector space $V$ is the quotient of the tensor product $bigotimes^k(V)$ by the relation $votimes v$.
To be somewhat more concrete and less abstract, this just means that for any vector $vin V$ the wedge product $vwedge v=0inbigwedge^2(V)$. Whenever you wedge vectors together, the result equals zero if at least two of the factors are linearly dependent.
Think of what happens to the cross product in $mathbb{R}^3$.

In fact, let $e_1,e_2,ldots,e_n$ be a basis of an inner product space $V$. Then $e_{i_1}wedge e_{i_2}wedge ldots wedge e_{i_k}$ is a basis for $bigwedge^k(V)$ where $1leq i_1 < i_2 < ldots < i_kleq n$.

If $V=mathbb{R}^3$ then $v wedge w$ equals $v times w$ up to signs of the entries. This seems a bit obscure because technically $vwedge w$ should be an element of $bigwedge^2(mathbb{R}^3)$. However, the latter vector space is isomorphic to $mathbb{R}^3$. In fact, this relation is true for all exterior powers given an orientation on the vector space.
The isomorphism is called the Hodge star operator.
It says that there is an isomorphism $starcolonbigwedge^{n-k}(V)tobigwedge^{k}(V)$. This map operators on a $(n-k)$-wedge $beta$ via the relation
$$
alpha wedge beta = langle alpha,starbeta rangle ,omega
$$
where $alphainbigwedge^{k}(V)$, $omega$ is an orientation form on $V$ and $langle cdot,cdot rangle$ is the induced inner product on $bigwedge^{k}(V)$ (see wiki). Notice that the wiki-page defines the relation the other way around.

How does all this answer your question you ask?
Well, let us take $k=1$ and $V=mathbb{R}^n$. Then the Hodge star isomorphism identifies the spaces $bigwedge^{n-1}(mathbb{R}^n)$ and $bigwedge^{1}(mathbb{R}^n)=mathbb{R}^n$. This is good because you originally wanted to say something about orthogonality between a set of $n-1$ linearly indepedent vectors $v_1,v_2,ldots,v_{n-1}$ and their "cross product".
Now let us exactly do that and set $beta :=v_1 wedge v_2 wedge ldots wedge v_{n-1}inbigwedge^{n-1}(mathbb{R}^n)$. Then the image $starbeta = star(v_1 wedge v_2 wedge ldots wedge v_{n-1})$ is a regular vector in $mathbb{R}^n$ and the defining condition above implies for $alpha=v_iinmathbb{R}^n=bigwedge^{1}(mathbb{R}^n)$
$$
v_i wedge (v_1 wedge v_2 wedge ldots wedge v_{n-1}) = alpha wedge beta = langle alpha,starbeta rangle ,omega = langle v_i,starbeta rangle ,omega.
$$
However, the left hand side equals zero for $i=1,2,ldots,n-1$, so that the vector $starbeta$ is orthogonal to all vectors $v_1,v_2,ldots,v_{n-1}$ which is what you asked for. So you might want to define the cross product of $n-1$ vectors as $v_1 times v_2 times ldots times v_{n-1} := star(v_1 wedge v_2 wedge ldots wedge v_{n-1})$.

Maybe keep in mind that the other two answers implicitly use the Hodge star operation (and also a basis) to compute the "cross product in higher dimension" through the formal determinant which is encoded in the use of the wedge product here.

You can work out the cross product $p$ in $n$-dimensions using the following:

$$p=detleft(begin{array}{lllll}e_1&x_1&y_1&cdots&z_1\e_2&x_2&y_2&cdots&z_2\vdots&vdots&vdots&ddots&vdots\e_n&x_n&y_n&cdots&z_nend{array}right),$$
where $det$ is the formal determinant of the matrix, the $e_i$ are the base vectors (e.g. $hat{i},hat{j},hat{k}$, etc), and $x,y,ldots,z$ are the $n-1$ vectors you wish to "cross".

You will find that $xcdot p=ycdot p=cdots=zcdot p=0$.

It's wonderful the determinant produces a vector with this property.

Yes, and apart from other answers an interesting approach to think about it is using Clifford's algebra.

This can introduce you the basic concept in a nonrigorous but approachable manner.

https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-visual-introduction/