题目内容
9.已知$|{\overrightarrow{TM}}|=2$,$|{\overrightarrow{TN}}|=4$,且$\overrightarrow{TM}•\overrightarrow{TN}=\frac{5}{2}$,若点P满足$|{\overrightarrow{TM}+\overrightarrow{TN}-\overrightarrow{TP}}|=2$,则$|{\overrightarrow{TP}}|$的取值范围为[3,7].分析 先根据向量的模的计算求出|$\overrightarrow{TM}$+$\overrightarrow{TN}$|=5,再根据绝对值的不等式得到|5-|$\overrightarrow{TP}$||≤2,解得即可
解答 解:∵$|{\overrightarrow{TM}}|=2$,$|{\overrightarrow{TN}}|=4$,且$\overrightarrow{TM}•\overrightarrow{TN}=\frac{5}{2}$,
∴|$\overrightarrow{TM}$+$\overrightarrow{TN}$|2=${\overrightarrow{TM}}^{2}$+${\overrightarrow{TN}}^{2}$+2$\overrightarrow{TM}$•$\overrightarrow{TN}$=4+16+5=25,
∴|$\overrightarrow{TM}$+$\overrightarrow{TN}$|=5
∵$|{|{\overrightarrow{TM}+\overrightarrow{TN}}|-|{\overrightarrow{TP}}|}|≤|{\overrightarrow{TM}+\overrightarrow{TN}-\overrightarrow{TP}}|$=2,
∴|5-|$\overrightarrow{TP}$||≤2,
∴5-2≤|$\overrightarrow{TP}$||≤5+2,
即3≤|$\overrightarrow{TP}$||≤7,
故答案为:[3,7]
点评 本题考查了向量的模的计算和绝对值不等式,属于中档题
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