Example 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) uncertain
(2006, National High School Mathematics Competition)
【Analysis】Convert the vector relationship of the condition into geometric characteristics.
Solution Let $t \overrightarrow{B C}=\overrightarrow{B D}$.
Then point $D$ lies on the line
$B C$, as shown in Figure 1. Therefore,
$$
\begin{aligned}
\overrightarrow{B A}-t \overrightarrow{B C} \\
=\overrightarrow{B A}-\overrightarrow{B D} \\
=\overrightarrow{D A} .
\end{aligned}
$$

Since $t \in \mathbf{R}$, hence
when $D$ traverses all points on the line $B C$, we have
$$
|\overrightarrow{D A}| \geqslant|\overrightarrow{A C}| \text {. }
$$

Therefore, $A C \perp B C$ at point $C$, i.e., $\triangle A B C$ is definitely a right triangle.
Hence, the correct choice is (C).