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有關向量的3條問題
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1. [此題另一先決條件是a 與b不平行] 若(a + b)與(a – b)平行, 則(a + b) = k(a – b)其中k為常數 (a + b) = k(a – b) a + b = ka – kb (k – 1)a = (k + 1)b 因a 與b不平行,所以a = 0; b= 0與已知條件矛盾 => (a + b) 與(a – b)不平行 2. (a + b).(a – b) = a.a – a.b + b.a – b.b = a.a – a.b + a.b – b.b = | a |2 - | b |2 3. AD = AB + BC + CD = (6,1) + (x,y) + (-2,-3) = (4 + x, y – 2) DA = -AD = (-4 – x, 2 – y) 向量BC//向量DA => BC = kDA (x, y ) = k(-4 – x, 2 – y) (x, y) = (-4k – kx, 2k – ky) x = -4k – kx … (1) y = 2k – ky … (2) (1) => k = -x / (4 + x) Sub into (2), y = k(2 – y) y = -x(2 – y)/(4 + x) y(4 + x) = -x(2 – y) 4y + xy = -2x + xy 2x + 4y = 0 x + 2y = 0
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有關向量的3條問題
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1. 已經 向量a 不等於 0 , 向量b 不等於 0. 求証: 向量a+向量b 與 向量a-向量b 不平行 2. 証明 : (向量a+向量b) ．(向量a-向量b) = 向量a^2 -向量b^2 3.如圖, 向量AB=(6,1) , 向量BC=(x,y) , 向量CD=(-2,-3) , 且向量BC//向量DA , 試確定 x , y 的關係式. 圖：http://i595.photobucket.com/albums/tt39/faifai32/a.jpg 答案: x+2y=0 求過程... 唔該晒~~ 更新: 第一題 唔係「已經」, 係「已知」最佳解答:
1. [此題另一先決條件是a 與b不平行] 若(a + b)與(a – b)平行, 則(a + b) = k(a – b)其中k為常數 (a + b) = k(a – b) a + b = ka – kb (k – 1)a = (k + 1)b 因a 與b不平行,所以a = 0; b= 0與已知條件矛盾 => (a + b) 與(a – b)不平行 2. (a + b).(a – b) = a.a – a.b + b.a – b.b = a.a – a.b + a.b – b.b = | a |2 - | b |2 3. AD = AB + BC + CD = (6,1) + (x,y) + (-2,-3) = (4 + x, y – 2) DA = -AD = (-4 – x, 2 – y) 向量BC//向量DA => BC = kDA (x, y ) = k(-4 – x, 2 – y) (x, y) = (-4k – kx, 2k – ky) x = -4k – kx … (1) y = 2k – ky … (2) (1) => k = -x / (4 + x) Sub into (2), y = k(2 – y) y = -x(2 – y)/(4 + x) y(4 + x) = -x(2 – y) 4y + xy = -2x + xy 2x + 4y = 0 x + 2y = 0
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