Consider the two regular submanifolds $A=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2-z^2=1\}$ $B=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2+z^2=1\}$of $\mathbb{R}^3$. I want to check if the intersect transverally, i.e, I want to check whether $\forall p\in A\cap B$, $T_pA+T_pB=T_p{\mathbb{R^3}}$. After some work, $$T_{(a,b,c)}A=\{(x,y,z)|ax+by-cz=0\}$$ $$T_{(a,b,c)}B=\{(x,y,z)|ax+by+cz=0\}$$
However, I do not know how to proceed from here onwards. Any help is appreciated.
The intersection is the circle $x^2+y^2=1,$ $z=0.$ For any point $p=(a,b,0)$ in the intersection you have
$$T_{(a,b,0)}A= T_{(a,b,0)}B=\{(x,y,z)\text{ | }ax+by=0\}.$$
Note that $T_x\mathbf{R}^3=\mathbf{R}^3$ for any $x \in \mathbf{R}^3.$ Thus the equation $T_{(a,b,0)}A+ T_{(a,b,0)}B=T_x\mathbf{R}^3=\mathbf{R}^3$ never holds for $p=(a,b,0)$ in the intersection (as $T_{(a,b,0)}A=T_{(a,b,0)}B$ is a 2-dimensional vector space), and hence the intersection is not transversal.