If you have 3 orthonormal vectors, $\pmatrix{u_1 \\ u_2 \\ u_3}$, $\pmatrix{v_1 \\ v_2 \\ v_3}$, $\pmatrix{w_1 \\ w_2 \\ w_3}$ such that $u= v \times w$ (crossproduct) can it be true that $v_i v_j + w_i w_j= - u_i u_j$?
I also get that $w_i ^2 + v_i ^3 = 1- u_i ^2$. How do you explain this? I am working with some product of vectors and I get that this must be true?
Combining them into a matrix $$A=\begin{bmatrix} u_1 & v_1 & w_1\\ u_2 & v_2 & w_2\\ u_3 & v_3 & w_3 \end{bmatrix}$$ then what you've observed can be restated as:
$$(\textsf{row }i)\cdot(\textsf{row }j)=\begin{cases} 1 & \text{if }i=j,\\ 0 & \text{if }i\neq j \end{cases}$$ In other words, you've observed that if we make a square matrix $A$ with orthonormal columns, then the rows of $A$ are also orthonormal to each other.
The rows of $A$ are just the columns of the transpose matrix, denoted $A^{\mathsf{T}}$.
A square matrix whose columns are orthonormal to each other is called an "orthogonal matrix" (poor naming choice, unfortunately). See here on Wikipedia.
Thus, the final restatement of what you've observed is that the transpose of an orthogonal matrix is again an orthogonal matrix.
How to prove this observation?
Note that because the columns of $A$ are orthonormal, we have $$A^{\mathsf{T}}A=\begin{pmatrix} u\cdot u & u\cdot v & u\cdot w\\ v\cdot u & v\cdot v & v\cdot w\\ w\cdot u & w\cdot v & w\cdot w \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=I$$ It is an important fact that for any square matrices $A$ and $B$, if we have $BA=I$, then we must also have $AB=I$. There is a stack exchange thread about that here.
But now, observe that because $A^{\mathsf{T}}A=I$, we must have $AA^{\mathsf{T}}=I$, so that $$AA^{\mathsf{T}}=\begin{pmatrix} (\textsf{row }1)\cdot(\textsf{row }1) & (\textsf{row }1)\cdot(\textsf{row }2) & (\textsf{row }1)\cdot(\textsf{row }3)\\ (\textsf{row }2)\cdot(\textsf{row }1) & (\textsf{row }2)\cdot(\textsf{row }2) & (\textsf{row }2)\cdot(\textsf{row }3)\\ (\textsf{row }3)\cdot(\textsf{row }1) & (\textsf{row }3)\cdot(\textsf{row }2) & (\textsf{row }3)\cdot(\textsf{row }3) \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=I$$ Thus the rows of $A$ are also orthonormal to each other.
