MT325垂心的向量形式

Posted 2021-02-15 mathstudy

设$H$为垂心,且$3overrightarrow{HA}+4overrightarrow {HB}+5overrightarrow {HC}=overrightarrow 0$,则$cosangle AHB=$____

分析:$ an Aoverrightarrow {HA}+ an Boverrightarrow {HB}+ an Coverrightarrow {HC}=overrightarrow 0$,
故$tan A:tan B:tan C=3:4:5$ 又$ an A an B an C= an A an B an C $
故$( an A, an B, an C)=(dfrac{3}{sqrt{5}},dfrac{4}{sqrt{5}},sqrt{5})$, 
从而$cosangle{AHB}=-cos C=-dfrac{sqrt{6}}{6}$

练习:$Delta ABC $中$AB=4,AC=3,BC=2$,点$H$为三角形的垂心,
若$overrightarrow {AH}=xoverrightarrow {AB}+yoverrightarrow {AC}$则$dfrac{y}{x}=$_____
答案:$-dfrac{11}{3}$
另外不常用的一个外心的结论$sin 2A overrightarrow {OA}+sin2Boverrightarrow {OB}+sin2Coverrightarrow {OC}=overrightarrow 0$