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# 兩題有關高中向量的問題

1.在三角形ABC中,若角BAC的角平分線交BC於D,則BD:DC=AB:AC

2.在三角形ABC中,AB=AC=4 BC=6 P 為外心

(1) 求AP跟AB的內積 ANS:8

(2)設AP向量=xAB向量+yAC向量 x.y為何? ans: x=y=4/7

### 1 個解答

• ?
Lv 7
6 年前
最佳解答

1.ΔABC: ∠BAC的角平分線交BC於D, BD:DC=AB:AC作AH⊥BC交BC於H點∠BAD = ∠DAC = QΔABD = 0.5*AB*AD*sinQ = 0.5*BD*AHΔACD = 0.5*AC*AD*sinQ = 0.5*CD*AH相除: AB/AC = BD/CD

2.c=AB=8, a=BC=7, b=CA=6, I=ΔABC內心，[ ]=向量, [AI]=r*[AB]+s*[AC]r.s=? Ans: r=2/7 s=8/21

作AH⊥BC交BC於H點: BH = x, HC = 7 - xy^2 = AH^2= 64 - x^2= 36 - (7-x)^2=> x = (64 + 49 - 36)/14 = 11/2=> y = √(64 - 5.5^2) = (√15)*3/2 => B = (0,0) => A = (11/2,3√15/2)=> C = (6,0)2s = a + b + c = 7 + 6 + 8 = 21

I = (a*A + b*B + c*C)/2s= [7*(11/2,3√15/2) + 6*(0,0) + 8*(7,0)]/21 = (9/2, (√15)/2)

[AI] = {b*[AB] + c*[AC]}/2s=> r = b/2s = 6/21 = 3/7=> s = c/2s = 8/21

3.ΔABC: AB=AC=4, BC=6, P=外心(1) [AP].[AB] = ? Ans = 8PD ⊥ BC, BD = DCPF ⊥ AB, AF = FBB = (0,0), D = (3,0), C = (6,0)A = (3,√(16-9)) = (3,√7)F = A/2 = (3/2,√7/2)Line PD: x = 3Line AB: y = √7*x/3 Line FD: (y - √7/2)/(x - 3/2) = -3/√7Lines PD & PF: y = -1/√7=> P = (3,-1/√7)[AP] = P - A= (3,-1/√7) - (3,√7)= (0, -8/√7)[AB] = B - A = (-3,-√7)

=> [AP].[AB] = 3*0 + (8/√7)*√7 = 8

(2) [AP] = x*[AB]+y*[AC]; x.y = ? Ans: x=y=4[AC] = C - A = (6,0) - (3,√7)= (3,-√7)

=> (0,-8/√7) = x*(-3,-√7) + y*(3,-√7)= (3(-x+y), -√7(x+y))=> -x + y = 0, x + y = 8/7=> x = y = 4/7Textbook answer = 4 = Error

2015-03-27 18:14:34 補充：

取消 Textbook answer = 4 = Error