Let $K_i: \mathbb{R}\mapsto \mathbb{R}^k$ are continuous functions for all $i=1,\dots,k-d$ such that for every fixed $t\in\mathbb{R}$ we have ${\cal K}_t=\{K_1(t),\dots,K_{k-d}(t)\}$ be a linear independent set in $\mathbb{R}^k$.
I'd like to have $d-$vector functions $B_1(),\dots,B_d()$ such that $B_i$ are continuous, $B_i(t)\bot K_j(t)$ for all $t$, for every $i=1,\dots,d$ and $j=1,\dots,k-d$, and also \begin{equation} {\cal K}_t\cup\{B_1(t),\dots,B_d(t)\} \end{equation} is a basis for $\mathbb{R}^k$.
To do this, note that from the standard basis $\{e_1,\dots,e_k\}$, we can choose $d-$vectors of them called by $f_1,\dots,f_d$ such that ${\cal K}_t\cup\{f_1,\dots,f_d\}$ is basis. But, how can we make they are orthogonal with $\cal K_t$. As I know, the gram schmidt produces an orthogonal basis, but I need to have partly orthogonal by fixing the $(k-d)-$vectors in my basis. Any suggestion? Thanks for any help.