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  • MT【325】垂心的向量形式

    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$sin 2A overrightarrow {OA}+sin2Boverrightarrow {OB}+sin2Coverrightarrow {OC}=overrightarrow 0$

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  • 原文地址:https://www.cnblogs.com/mathstudy/p/10674736.html
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