Given that:
$x_1 x_2 + y_1 y_2 + z_1 z_2=0$,
$x_1 x_3 + y_1 y_3 + z_1 z_3=0$,
$x_2 x_3 + y_2 y_3 + z_2 z_3=0$,
$x_1^2+y_1^2+z_1^2=1$,
$x_2^2+y_2^2+z_2^2=1$,
$x_3^2+y_3^2+z_3^2=1$
How to prove the following:
$x_1^2+x_2^2+x_3^2=1$,
$x_1 y_1 + x_2 y_2 + x_3 y_3=0$
With Linear Algebra, this is straightforward.
Let $U$ be the matrix whose rows are the 3 vectors $(x_i,y_i,z_i)$. Then, you’re given that $U U^T=I$. Therefore $U$ is invertible with inverse $U^T$. Since right inverses are left inverses, we also have $U^T U=I$ which simplifies to your desired equations.
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How can we do this without directly relying on the concept of invertibility in Linear algebra?
I’m still going to use Linear algebra to simplify the computations, but you could imagine writing this all out as linear equations of the underlying 9 variables.
We have that $|U||U^T|=1$, so $|U| \neq 0$.
Let $\mathbb{adj}(A)$ be the adjugate matrix defined as the transpose of the matrix of cofactors. If you expanded this out, then by the cofactor definition of determinant, it satisfies $A \mspace4mu\mathbb{adj}(A) = \mathbb{adj}(A) A=|A| I$. This is just a simple equality if your write it out in terms of the underlying 9 variables.
Let’s try simplifying $\mathbb{adj}(U) U U^T U$.
Collapsing from the left side, we have: $\mathbb{adj}(U) U U^T U=|U|I U^T U= |U| U^T U$
Collapsing the middle first, we have: $$\mathbb{adj}(U) U U^T U=\mathbb{adj}(U) I U = \mathbb{adj}(U) U=|U| I$$
Thus, $$|U| U^T U =\mathbb{adj}(U) U U^T U= |U| I$$
Since $|U|\neq 0$, we can divide by it:
$$U^T U=I$$
Expanding this gives the desired result.