6. Given vectors $\overrightarrow{O A}, \overrightarrow{O B}$ are perpendicular, $|\overrightarrow{O A}|=|\overrightarrow{O B}|=24$. If $t \in[0,1]$, then the minimum value of $|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right|$ is
A. $2 \sqrt{193}$
B. 26
C. $24 \sqrt{2}$
D. 24
Let $\overrightarrow{O A}=(24,0), \overrightarrow{O B}=(0,24)$,
then $|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right|=|t \overrightarrow{O B}+(1-t) \overrightarrow{O A}|+\left|\left(\frac{7}{12}-t\right) \overrightarrow{O B}-(1-t) \overrightarrow{O A}\right|$ $=24\left[\sqrt{t^{2}+(1-t)^{2}}+\sqrt{\left(\frac{7}{12}-t\right)^{2}+(1-t)^{2}}\right]$.
The above expression represents 24 times the sum of the distances from the point $(t, 1-t)$ to the points $(0,0)$ and $\left(\frac{7}{12}, 0\right)$. By reflecting the point $(0,0)$ over the line $y=1-x$ to get the point $(1,1)$, we know that the minimum value of the above expression is $24 \sqrt{\left(\frac{5}{12}\right)^{2}+1}=26$, so the answer is $B$.