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2024届陕西省九年级学业水平质量监测(♪)数学h试卷答案

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分析利用向量模的关系，建立坐标系，求出相关点的坐标，分别求出直线CE，DB的方程，求出交点即G点的坐标，然后求解向量的模即可．

解答解：以A点为原点，以$\overrightarrow{AB}$为x轴，建立如图所示的坐标系，
∵|$\overrightarrow{AB}$|=8，|$\overrightarrow{AC}$|=6，∠BAC=$\frac{π}{3}$，$\overrightarrow{AD}$=$\overrightarrow{DB}$，$\overrightarrow{AE}$=2$\overrightarrow{EC}$，
∴A（0，0），B（8，0），E（4，0），D（2，2$\sqrt{3}$），C（3，3$\sqrt{3}$），
∴直线CE的方程为$\frac{y-0}{3\sqrt{3}-0}$=$\frac{x-4}{3-4}$，即3$\sqrt{3}$x+y-12$\sqrt{3}$=0，①
直线DB的方程为$\frac{y-0}{2\sqrt{3}-0}$=$\frac{x-8}{2-8}$，即x+$\sqrt{3}$y-8=0，②
由①②构成方程组，解得$\left\{\begin{array}{l}{x=\frac{7}{2}}\\{y=\frac{3\sqrt{3}}{2}}\end{array}\right.$，
∴点G（$\frac{7}{2}$，$\frac{3\sqrt{3}}{2}$），
∴$\overrightarrow{AG}$=（$\frac{7}{2}$，$\frac{3\sqrt{3}}{2}$），
∴|$\overrightarrow{AG}$|=$\sqrt{（\frac{7}{2}）^{2}+（\frac{3\sqrt{3}}{2}）^{2}}$=$\sqrt{19}$，
故选：B

点评本题考查向量的几何中的应用，向量的坐标运算，直线方程，直线的交点，向量的模，考查计算能力，属于中档题．

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