已知 $\vec{a}=(1,2, \sqrt{3}), \vec{b}=(-1, \sqrt{3}, 0)$, 则 $\vec{a} \cdot \vec{b}+|\vec{b}|=$
because \vec{a}=(1,2, \sqrt{3}), \vec{b}=(-1, \sqrt{3}, 0) \quad 	herefore \vec{a} \cdot \vec{b}=-1+2 \sqrt{3}$,$|\overrightarrow{\mathrm{b}}|=\sqrt{1+3}=2, \quad 	herefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}+1$,