1. Given $\triangle A B C$. If for any $t \in \mathbf{R}, \mid B A- t B C|\geqslant| A C \mid$, then $\triangle A B C(\quad)$.
(A) must be an acute triangle
(B) must be an obtuse triangle
(C) must be a right triangle
(D) the answer is uncertain
-1.C.
Let $\angle ABC = \alpha$, and draw $AD \perp BC$ at point $D$.
From $|BA - tBC| \geq |AC|$, we get
$|BA|^2 - 2tBA \cdot BC + t^2|BC|^2 \geq |AC|^2$.
Let $t = \frac{BA \cdot BC}{|BC|^2}$, substituting into the above equation gives
$|BA|^2 - 2|BA|^2 \cos^2 \alpha + \cos^2 \alpha |BA|^2 \geq |AC|^2$,
which simplifies to $|BA|^2 \sin^2 \alpha \geq |AC|^2$, or $|BA| \sin \alpha \geq |AC|$.
Thus, we have $|AD| \geq |AC|$.
Therefore, $\angle ACB = \frac{\pi}{2}$.