F.5_AMAHS_Vector

唔知點解我換唔到行，所以唯有用『(換行)』代替~
唔好意思... XD
(換行)
13:
(換行)
Construct two lines from C perpendicular to u and v.
(換行)
The result below follows immediately:
(換行)
AC = (10cos40)v + (10cos25)u
　 = 7.66v + 9.06u [corr. to 3 s.f.]

(換2行~)

14:
(換行)

|u| = sqrt ( (1/2)^2 + (sqrt3 / 2)^2) {<--sqrt = square root.}

　= 1
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Therefore u is a unit vector.
(換行)

The vertex of u is ( sqrt3/ 2 , 1/2 ).
(換行)
Then the required line passes through ( 0 , 0 )and ( sqrt3/ 2 , 1/2 ).
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i.e. The equation is y = x/sqrt3.

(換行)
The line perpendicular to u is y = -sqrt3 x
(換行)
The unit vector perpendicular to u is v = -1/2 i + sqrt3/2 j

(換行)
a = 5i + 5sqrt3 j

　= 10 ( -1/2i + sqrt3 / 2 j)

= 10 v

2009-01-04 22:25:35 補充：
原來預覽先至換唔到~~~ xd
不便之處敬請原諒~~~ XDXDXDXD