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發問:
@@@@@@@@@@@@@@@@@@@@@@@2 圖片參考：http://img546.imageshack.us/img546/5375/30526846.jpg
最佳解答:
a) AB = AD + DB = -2a + 6 b DM = (DB + DA) / 2 = (6b + 2a)/2 = 3b + a b) i) BC = kDM BC = k(3b + a) = ka + 3kb AC = 2DB AC = 2(6b) = 12b AB = -2a + 6b AB + BC = AC -2a + 6b + ka + 3kb = 12b ka + 3kb = 2a + 6b k(a + 3b) = 2(a + 3b) k = 2 ii) BC = 2DM Thus BC // DM, i.e. BC // DN. And AM = MB By intercept theorem, AN = (1/2)AC
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Vectors發問:
@@@@@@@@@@@@@@@@@@@@@@@2 圖片參考：http://img546.imageshack.us/img546/5375/30526846.jpg
最佳解答:
a) AB = AD + DB = -2a + 6 b DM = (DB + DA) / 2 = (6b + 2a)/2 = 3b + a b) i) BC = kDM BC = k(3b + a) = ka + 3kb AC = 2DB AC = 2(6b) = 12b AB = -2a + 6b AB + BC = AC -2a + 6b + ka + 3kb = 12b ka + 3kb = 2a + 6b k(a + 3b) = 2(a + 3b) k = 2 ii) BC = 2DM Thus BC // DM, i.e. BC // DN. And AM = MB By intercept theorem, AN = (1/2)AC
其他解答:
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