Let $V \subseteq \mathbb{R}^3$ be a two-dimensional vector space, and suppose we have a family of isometric embeddings $A_t:V \to \mathbb{R}^3$,
in the sense that $A_t$ is a linear isometric embedding of $V$ (endowed with the inner product induced by the Euclidean on on $\mathbb{R}^3$) into $\mathbb{R}^3$.
It is easy to see that for each $t$, there exist a unique isometric embedding $q_t :V^{\perp} \to \mathbb{R}^3$, such that $A_t \oplus q_t:\mathbb{R}^3 \to \mathbb{R}^3 $ is an isometry.
My question is wether the regularity of $q_t$ "equals" to the regularity of $A_t$, that is suppose $t \to A_t$ is differentiable (or $C^k$). Is it true that $t \to q_t$ is also differentiable (or $C^k$)?
You didn't say exactly what you mean by "smoothness" of the maps $t\mapsto A_t$ and $t\mapsto q_t$. I'm going to assume that whatever definition you have in mind is equivalent to the following: If you choose a fixed orthonormal basis $(b_1,b_2)$ for $V$, then the vectors $A_t(b_1)$ and $A_t(b_2)$ depend smoothly on $t$; and similarly for $q_t$.
With this interpretation, the answer is yes: For each $t$, the vectors $A_t(b_1)$ and $A_t(b_2)$ form an orthonormal basis for $A_t(V)$, and then their cross-product $A_t(b_1)\times A_t(b_2)$ is an (ortho)normal basis for $(A_t(V))^\perp$. So we can define $q_t$ by letting $b_3 = b_1\times b_2$ (which is an (ortho)normal basis for $V^\perp$), and setting $q_t (b_3) = A_t(b_1)\times A_t(b_2)$, which also depends smoothly on $t$.