Prove that the following set is a manifold and find it's dimension $$M=\{(x,y)\in\Bbb{R^3}\times\Bbb{R^3}|x\cdot y=0, \|x\|=\|y\|=1\}$$
So $M$ can be thought of as the intersection of the following equations: $$x_1y_1+x_2y_2+x_3y_3=0$$ $$x_1^2+x_2^2+x_3^2=1$$$$y_1^2+y_2^2+y_3^2=1$$
I tried defining an implicit function $F(x,y)$ s.t $F(x,y)=0$ iff $(x,y)\in M$ but didn't know how to take it from here. Any help would be appreciated.
What if you consider the $\mathcal C^1$ function $\mathbf F : \mathbb R^6 \to \mathbb R^3$ defined as: $$\mathbf F(\mathbf x)= \begin{bmatrix} x_1y_1+x_2y_2+x_3y_3 \\ x_1^2+x_2^2+x_3^2 -1 \\ y_1^2+y_2^2+y_3^2-1 \\ \end{bmatrix} $$
Can you check $\text{rank}(D \mathbf F(\mathbf x))$ now?