Tensor in D-dimensional space crosswise with 2 vectors

[olgerm]

I have 2 vectors $\vec {V}=(v_1,v_2,v_3,v_4...v_D)$ and $\vec {X}=(x_1,x_2,x_3,x_4...x_D)$ in D-dimensional euclidean space.
I want a tensor ,which is crosswise with both of them.

I think that the tensor is parallel with (D-2)-dimensional area, am I right?
I do not know a lot about tensors ,so in answer please also explain me the notation that you used.

 

[fzero]

It's not exactly clear what you want. Do you mean a rank 2 tensor, which we could think of as a matrix? In that case, we can have two possible conditions:

1. $\sum_j M_{ij} v_j =0$, or
2. $\sum_j v_j M_{ji}=0$.

You might also want both of these to be satisfied. These conditions are independent unless $M$ is symmetric, $M_{ij}=M_{ji}$.

Either way, an analog of the Gram-Schmidt process allows you to start with any tensor $M$ and construct a new tensor $M'$ that satisfies one or both of these conditions. For instance, as long as $|V|\neq 0$,
$$ M'_{ij} = M_{ij} - \sum_k M_{ik} v_k \frac{v_j}{|V|^2}$$
will satisfy $\sum_j M'_{ij} v_j = 0$. The other combinations should be easy to work out.

Hopefully this component notation is familiar, or at least obvious to you.

 

[olgerm]

Do you mean a rank 2 tensor, which we could think of as a matrix?

I can´t explain ,what I mean ,in terms of tensor ,but I´ll try to explain it in terms of what I want it for:
There are 4 vectors $\vec V_1$ ,$\vec V_2$ ,$\vec X$ and $\vec F$.
$F⊥V_2$
F is parallel with plane ,which is also parallel with vectors $\vec X$ and $\vec V_1$

I know only value of vectors $\vec X$ and $\vec V_1$. My friend knows only value of vector $\vec V_2$. I want a tensor that I could give to my friend so he could calculate value of $\vec F$ without knowing vectors $\vec X$ and $\vec V_1$ separately.

 

[olgerm]

Nb: by value I meant the vector itself not length of vector.

I there any such tensor at all ,I wanted? If there is not, it would be useful to learn that there is not.