1. Given $\triangle A B C$, if for any $t \in \mathbf{R},|\overrightarrow{B A}-t \overrightarrow{B C}| \geqslant|\overrightarrow{A C}|$, then $\triangle A B C$ must be
A. Acute triangle
B. Obtuse triangle
C. Right triangle
D. The answer is uncertain
$$
\begin{array}{l}
|\overrightarrow{B A}-t \overrightarrow{B C}|^{2}=a^{2} t^{2}-2 c a t \cos B+c^{2}=(a t-c \cos B)^{2}+c^{2} \sin ^{2} B \\
\Rightarrow b \leqslant c \sin B \Rightarrow \sin B \leqslant \sin C \sin B \Rightarrow \sin C \geqslant 1 \Rightarrow C=\frac{\pi}{2} . \text { So the answer is } C .
\end{array}
$$