Question

在△ABC中AB=2,AC=3,∠A=60°P是三角形的内心,求向量AP*向量BC

Answer 1

设AP延长线交BC于D,连接BP,则 |AB|/|AC|=|BD|/|DC|=2/3 ===>BD=2/5BC=2/5(AC-AB)

AD=AB+BD=AB+2/5BC=AB+2/5(AC-AB)=3/5AB+2/5AC

所以 ｜BD|^2=4/25[AC-AB]^2=4/25(AC^2-2AC*AB+AB^2)=4/25(3^2-2*3*2*COS60+2^2)=28/25

|BD|=2根号7/5

| AP|/|PD|=|AP|/(|AD|-|AP|)=|AB|/|BD|=2/(2根号7/5)=5/根号7===》｜AP|/|AD|=5/(5+根号7）===》AP=5/(5+根号7)AD=5/(5+根号7)(3/5AB+2/5AC)=1/(5+根号7)*(3AB+2AC)

所以 向量AP*向量BC= 1/(5+根号7)*(3AB+2AC)*(AC-AB)=1/(5+根号7)*(3AB*AC-3AB^2+2AC^2-2AB*AC)=1/(5+根号7)*(2*3^2-3*2^2+AB*AC)=1/(5+根号7)*(6+2*3*COS60)=1/(5+根号7)*9

=(5-根号7)/2.