1. In $\triangle A B C$, $A B=3, A C=5, \overrightarrow{B A} \cdot \overrightarrow{B C}=1$, then $B C=$
A. $2 \sqrt{6}$
B. $\sqrt{15}$
C. $3 \sqrt{2}$
D. $\sqrt{17}$
$$
\begin{array}{l}
\overrightarrow{B C}^{2}=(\overrightarrow{A C}-\overrightarrow{A B})^{2}=25+9-2 \overrightarrow{A C} \cdot \overrightarrow{A B}=34-2(\overrightarrow{A B}+\overrightarrow{B C}) \cdot \overrightarrow{A B} \\
=34-2 \overrightarrow{A B}^{2}-2 \overrightarrow{B C} \cdot \overrightarrow{A B}=34-18+2=18 \Rightarrow B C=3 \sqrt{2} \text {, so the answer is } C .
\end{array}
$$