A Vector Question 1

圖片參考：http://i150.photobucket.com/albums/s87/mllokloks/vector2222.jpg

(a) OD=4i+3j
(b) let OB=xi+yj
then 4=kx/(k+1), 3=ky/(k+1)
x=4(k+1)/k ,y =3(k+1)/k
OB=[4(k+1)/k]i+[3(k+1)/k]j
AB=OB-OA=[4(k+1)/k]i+[3(k+1)/k]j-(6i+2j)
AB=[4(k+1)/k-6]i+[3(k+1)/k-2]j
(c)
AB.OB=0
[4(k+1)/k-6][4(k+1)/k]+[3(k+1)/k-2][3(k+1)/k]=0
[4(k+1)-6k][4(k+1)]+[3(k+1)-2k][3(k+1)]=0
(1-2k)(4k+4)+(k+1)(3k+3)=0
-8k^2-4k+4+3k^2+6k+3=0
5k^2-2k-7=0
(k+1)(5k-7)=0
k=-1 or 7/5
So k=7/5
(d)
BD : CD =1: √5
AD=CD
OB=[(7/5)+1]BD=(12/5)BD
OD=OB-BD=(7/5)BD
AD:OD=CD:OD=√5BD:(7/5)BD=5√5:7
(ii)
OC=2i+4j, OD=4i+3j
CD=OD-OC=2i-j
OC.CD=2*2-4*1=0
angle OCD= angle ABD=90
angle ODC= angle ADB
angle COD= angle BAD
So triangle OCD similar to triangle ABD