Vector Question no.3

by formula
OC
=1/(k+1)[kOA+OB]
=1/(k+1)[kau+bv]
(ii)
OB.OC=|OB||OC|cos(BOC)=b|OC|cos(BOC)
OA.OC=|OA||OC|cos(AOC)=a|OC|cos(AOC)
Since cos(AOC)=cos(BOC)
(OA.OC)/(a|OC|)=(OB.OC)/(b|OC|)
(OA.OC)/(a)=(OB.OC)/(b) [OC=1/(k+1)[kau+bv]]
[ka^2+ab(u.v)]/a=[kab(u.v)+b^2]/b
ka+b(u.v)=ka(u.v)+b
ka[1-(u.v)]=b[1-(u.v)]
k=b/a
(iii)
OA=6i+8j, OB=(15/2)j
a=10 , b=15/2
k=b/a=15/20=3/4
OC
=4/7[3/4(6i+8j)+(15/2)j]
=4/7[(9/2)i+(27/2)j]
=(18/7)i+(54/7)j



2007-02-19 21:14:41 補充：
PART (III)因為a和b其實即是OA和OB的長所以a=√(6^2+8^2)=10, b=15/2個j應該擺錯位