向量+垂心+面積

△ABC,AB=AC=5,BC=8,設H為△ABC垂心,向量AH=x向量AB+y向量Ac(1) (x,y)=?作AD⊥BC => BD=DC => AD = (AB + AC)/2C=(0,0), D=(-4,0), B=(-8,0) AB = 5, BD = 4 => A=(-4,3)ΔCDH: CH⊥BA => ∠HCD = 53 deg, HD = 4*tan(53) = 16/3=> H = (-4,16/3) = 跑到ΔABC外面=> AH = -[(AB + AC)/2]*|AH/AD|.....兩者反方向帶負號= -[(AB + AC)/2]*[(16/3-3)/3]= -[(AB + AC)/2]*(7/9)= -(AB + AC)*7/18=> ans=(-7/18,-7/18)
(2) △HAB之面積= ΔHBD - ΔABD= 0.5*4*(16/3 -3)= 2*7/3= 14/3