✔ 最佳答案
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
