题目内容
1.已知△ABC,AB=7,AC=8,BC=9,P为平面ABC内一点,满足$\overrightarrow{PA}•\overrightarrow{PC}$=-7,则|$\overrightarrow{PB}$|的最小值是4.分析 可考虑让条件$\overrightarrow{PA}•\overrightarrow{PC}=-7$中出现$\overrightarrow{PB}$:$(\overrightarrow{PB}+\overrightarrow{BA})•(\overrightarrow{PB}+\overrightarrow{BC})=-7$,进一步得到${\overrightarrow{PB}}^{2}+\overrightarrow{PB}•(\overrightarrow{BA}+\overrightarrow{BC})+\overrightarrow{BA}•\overrightarrow{BC}=-7$,可设$\overrightarrow{PB}$和$\overrightarrow{BA}+\overrightarrow{BC}$夹角为θ,从而得到$|\overrightarrow{PB}{|}^{2}+|\overrightarrow{PB}||\overrightarrow{BA}+\overrightarrow{BC}|cosθ+\overrightarrow{BA}•\overrightarrow{BC}=-7$,而根据条件可以求出$\overrightarrow{BA}•\overrightarrow{BC}$,$|\overrightarrow{BA}+\overrightarrow{BC}|$,这样根据-1≤cosθ≤1即可得出$|\overrightarrow{PB}|$的取值范围,从而得出$|\overrightarrow{PB}|$的最小值.
解答 解:根据条件:$\overrightarrow{BA}•\overrightarrow{BC}=|\overrightarrow{BA}||\overrightarrow{BC}|cosB$=$\frac{1}{2}({7}^{2}+{9}^{2}-{8}^{2})=33$;
∴$|\overrightarrow{BA}+\overrightarrow{BC}{|}^{2}={\overrightarrow{BA}}^{2}+2\overrightarrow{BA}•\overrightarrow{BC}+{\overrightarrow{BC}}^{2}$=49+66+81=196;
∴$|\overrightarrow{BA}+\overrightarrow{BC}|=14$;
由$\overrightarrow{PA}•\overrightarrow{PC}=-7$得$(\overrightarrow{PB}+\overrightarrow{BA})•(\overrightarrow{PB}+\overrightarrow{BC})$=${\overrightarrow{PB}}^{2}+\overrightarrow{PB}•(\overrightarrow{BA}+\overrightarrow{BC})+\overrightarrow{BA}•\overrightarrow{BC}=-7$;
∴${\overrightarrow{PB}}^{2}+|\overrightarrow{PB}||\overrightarrow{BA}+\overrightarrow{BC}|cosθ$$+\overrightarrow{BA}•\overrightarrow{BC}$=-7,θ为向量$\overrightarrow{PB}$和$\overrightarrow{BA}+\overrightarrow{BC}$的夹角;
∴${\overrightarrow{PB}}^{2}+14|\overrightarrow{PB}|cosθ+33=-7$;
∴$cosθ=-\frac{|\overrightarrow{PB}{|}^{2}+40}{14|\overrightarrow{PB}|}$,-1≤cosθ≤1;
∴$-1≤-\frac{|\overrightarrow{PB}{|}^{2}+40}{14|\overrightarrow{PB}|}≤1$;
解得$4≤|\overrightarrow{PB}|≤10$;
∴|$\overrightarrow{PB}$|的最小值为4.
故答案为:4.
点评 考查数量积的计算公式,余弦定理,向量长度的求法:$|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{(\overrightarrow{a}+\overrightarrow{b})^{2}}$,向量加法的几何意义,以及余弦函数的值域,解一元二次不等式.
A. | (-2,2) | B. | (-∞,2) | C. | (-∞,-2)∪(2,+∞) | D. | (2,+∞) |
A. | ($\overrightarrow{a}$•$\overrightarrow{b}$)•$\overrightarrow{c}$=$\overrightarrow{a}$•($\overrightarrow{b}$•$\overrightarrow{c}$) | B. | |$\overrightarrow{a}$-$\overrightarrow{b}$|2=|$\overrightarrow{a}$|2-2|$\overrightarrow{a}$||$\overrightarrow{b}$|+|$\overrightarrow{b}$|2 | ||
C. | 若|$\overrightarrow{a}$|=|$\overrightarrow{b}$|=|$\overrightarrow{a}$+$\overrightarrow{b}$|,则$\overrightarrow{a}$与$\overrightarrow{b}$的夹角为60° | D. | 若|$\overrightarrow{a}$|=|$\overrightarrow{b}$|=|$\overrightarrow{a}$-$\overrightarrow{b}$|,则$\overrightarrow{a}$与$\overrightarrow{b}$的夹角为60° |