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Vector~!!!
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As follows: 圖片參考：http://g.imagehost.org/0346/ScreenHunter_03_May_02_10_45.gif
最佳解答:
其他解答:
(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.
Vector~!!!
發問:
As follows: 圖片參考：http://g.imagehost.org/0346/ScreenHunter_03_May_02_10_45.gif
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
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°其他解答:
(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.
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