F.5_AMAHS_Vector

For u
cos25=(Au+Bv).u/10
9.063=A+Bu.v=A+0.4226B
Similarly
cos40=(Au+Bv).v/10
7.6604=0.4226A+B
So
7.6604=0.4226(9.063-0.4226B)+B
B=4.66
A=7.09
AC=7.09u+4.66v
14
u^2=3/2+1/2=1
So u is an unit vector
slope=(1/2)/(sqrt(3)/2)=1/sqrt(3)
So the line is y=[1/sqrt(3)]x
(b)
a.u=5sqrt(3)
|a|=10
So (a.u)^2+(the component of a alog a line perpendiculat to u)^2=10^2
=> the component of a alog a line perpendiculat to u=5

2009-01-12 00:17:33 補充：
u^2=3/4+1/4=1

2009-01-12 00:18:51 補充：
超凡學生﹐你用甚麼軟件做你的答案架﹐我想學呀

As follow AS~~~~

圖片參考：http://e.imagehost.org/0492/13_8.gif


圖片參考：http://e.imagehost.org/0549/14a.gif


圖片參考：http://e.imagehost.org/0030/14b.gif


2009-01-11 23:58:54 補充：
遲小.請給樓上最佳~!

2009-01-11 23:59:03 補充：
遲了.請給樓上最佳~!