6. Given that $\overrightarrow{O A} \perp \overrightarrow{O B}$, and $|\overrightarrow{O A}|=|\overrightarrow{O B}|=24$. If $t \in[0,1]$, then
$$
|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right|
$$

the minimum value is ( ).
(A) $2 \sqrt{193}$
(B) 26
(C) $24 \sqrt{2}$
(D) 24
6. B.

Construct a square $O A C B$, connect the diagonal $A B$, and let $D$ and $E$ be points on the diagonal $A B$ and the side $O B$ respectively, such that
$$
\begin{array}{l}
t \overrightarrow{A B}-\overrightarrow{A O}=\overrightarrow{O D}, \\
\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}=\overrightarrow{E D}, \\
E B=10, O D=D C . \\
\text { Therefore }|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right| \\
=|\overrightarrow{E D}|+|\overrightarrow{D C}| \leqslant|\overrightarrow{E C}|=26 .
\end{array}
$$