贵州省2024届六盘水市高三第三次诊断性监测数学试卷答案,目前收集并整理关于贵州省2024届六盘水市高三第三次诊断性监测数学试卷答案得系列试题及其答案，更多试题答案请关注我们网站

贵州省2024届六盘水市高三第三次诊断性监测数学试卷答案

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分析已知向量等式移向，平方求出$\overrightarrow{OA}•\overrightarrow{OB}=0$，由$3\overrightarrow{OA}+4\overrightarrow{OB}=5\overrightarrow{OC}$得出A，B，C三点在圆心的同一侧，从而得出圆周角∠C的大小；由AB=1求出$|\overrightarrow{OA}|=|\overrightarrow{OB}|=\frac{\sqrt{2}}{2}$，把$\overrightarrow{OC}$、$\overrightarrow{AB}$用$\overrightarrow{OA}、\overrightarrow{OB}$表示，展开后得答案．

解答解：∵3$\overrightarrow{OA}$+4$\overrightarrow{OB}$-5$\overrightarrow{OC}$=$\overrightarrow{0}$，
∴$3\overrightarrow{OA}+4\overrightarrow{OB}=5\overrightarrow{OC}$，
两边平方可得：$9|\overrightarrow{OA}{|}^{2}+24\overrightarrow{OA}•\overrightarrow{OB}$$+16|\overrightarrow{OB}{|}^{2}$=$25|\overrightarrow{OC}{|}^{2}$．
∵A，B，C在圆上，设OA=OB=OC=1．
∴$\overrightarrow{OA}•\overrightarrow{OB}=0$，
由$3\overrightarrow{OA}+4\overrightarrow{OB}=5\overrightarrow{OC}$，可知A，B，C三点在圆心的同一侧，
∴根据圆周角定理知∠C=180°-$\frac{1}{2}$90°=135°；
故答案为：135°；
若AB=1，则$|\overrightarrow{OA}|=|\overrightarrow{OB}|=\frac{\sqrt{2}}{2}$，
∴$\overrightarrow{OC}$•$\overrightarrow{AB}$=$（\frac{3}{5}\overrightarrow{OA}+\frac{4}{5}\overrightarrow{OB}）•（\overrightarrow{OB}-\overrightarrow{OA}）$=$\frac{1}{5}$．
故答案为：135°；$\frac{1}{5}$．

点评本题考查平面向量的数量积运算，考查三角形外心的性质和应用，解题的关键是对于所给的向量式的整理，注意向量运算法则的灵活运用，是中档题．

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