一些數學向量的題目

Question

一. 3AB=4BC-5DA 若BD=xAC yBA X=? Y=?
二.在四邊形ABCD中 角A=120度 AB=1 AD=2且AC=3AB 2AD球AC的長度?
希望可以用簡單一點的方法~太久沒碰~都忘光光了.........

Answer 1

1. 3AB = 4BC - 5DA 若
BD
= xAC＋ yBA
= x(AB＋BC) - yAB
= (x-y)AB＋xBC
= (x-y)AB＋x[(3/4)AB＋(5/4)DA)]
= (7x/4 - y)AB ＋(5x/4)DA ... (1)
又 BD = BA＋AD ... (2)
由 (1) (2) => 7x/4 - y = -1; 5x/4 = -1
=> x = -4/5, y = -2/5
2. AC = 3AB＋2AD
延長 AB 使得 AE = 3AB => |AE| = 3|AB| = 3
連 EC
則 AC = AE＋EC = 3AB＋EC
所以 EC = 2 AD
=> EC // AD, |EC| = 2|AD| = 4, ∠AEC = 180 - 120 = 60
餘弦定理
|AC|2 = |AE|2＋|EC|2 - 2|AE|*|EC|*cos(60)
= 9＋16 - 2*3*4*(1/2)
= 13
=> |AC| = √13
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.