Vector~!!!

a. 設 QR:PQ = r:s，則 OQ = [rOP+sOR]/[r+s]
於是有 x= r/[r+s], y= s/[r+s]，易得 x+y=1。

bi. 由題設知 AM = 3/5*AB
於是 AT = 5/3*m*AM + n*AC
由(a)得 5/3*m+n=1。

ii. 由題設知 AN = 1/3*AC
於是 AT = m*AM + 3n*AC
由(a)得 m+3n=1。

由上面結果，解得 m=1/2, n=1/6。

iii. 由 bii，有 AT = 1/2*AB+1/6*AC
於是有 AH = λ*AT = λ/2*AB+λ/6*AC

因為C、H、B共線，由(a)得 λ/2+λ/6 = 1
於是解得 λ= 3/2

ci. AT = 1/2*AB+1/6*AC = 1/2*(2i+j)+1/6*(i-2j) = 7/6*i+1/6*j
BC = AC - AB = (i-2j) - (2i+j) = -i-3j

ii. |AT| = 1/6*√[7^2+1^2] = 1/6*√50
|BC| = √[1^2+3^2] = √10
AT‧BC = -7/6 - 3/6 = -5/3

由 AT‧BC = |AT|*|BC|*cos∠AHC
得 -5/3 = 5/3*√5*cos∠AHC
cos∠AHC = -1/√5
∠AHC = 117度 或 63度 [視乎大於直角還是小於直角]


2009-05-04 20:45:46 補充：
由 AT‧BC = |AT|*|BC|*cos(180° - ∠AHC)
得 -5/3 = 5/3*√5*cos(180° - ∠AHC)
cos∠AHC = 1/√5
∠AHC = 63°

(a) Let OP = a and OR = b, so RP = a- b.
Let RQ = mRP = m( a - b).
OQ = OR + RQ = b + m(a - b) = b + ma - mb = ma + ( 1 - m)b = mOP + ( 1- m) OR
that means x = m and y = 1 - m, so x + y = 1.