兩題數學的向量的問題....

請參考
https://skydrive.live.com/redir.aspx?cid=380a4a49a37829dc&resid=380A4A49A37829DC!115&parid=380A4A49A37829DC!103&authkey=!APOVQ2G5nISdg8o

第2題可以取BD的法向量*根號3
之後中點加上(BD的法向量*根號3) 即為一解

1.
a=[Cosθ，Sin(θ+π/6)]
｜a｜^2=Cos^2 θ+Sin^2 (θ+π/6)
4｜a｜^2=4Cos^2 θ+4Sin^2 (θ+π/6)
=2+2Cos(2θ)+2－2Cos(2θ+π/3)
=4+2Cos(2θ)－2Cos(2θ)Cos(π/3)+2Sin(2θ)Sin(π/3)
=4+2Cos(2θ)－Cos(2θ)+√3Sin(2θ)
=4+Cos(2θ)+√3Sin(2θ)
=4+2[1/2*Cos(2θ)+√3/2*Sin(2θ)]
=4+2[Sin(π/6)*Cos(2θ)+Cos(π/6)*Sin(2θ)]
=4+2Sin(π/6+2θ)
0<=θ>=2π
0<=2θ>=4π
π/6<=2θ+π/6<25π/6
2θ+π/6=π/2 or 2θ+π/6=5π/2
θ=π/6 or θ=7π/6
when θ=π/6 or θ=7π/6
4|a|^2=4+2=6
｜a｜=√3/2

2
Sol
C(p，q)，AB中點D(1，0.5)
AB直線方程式：(y+1)/(x－3)=(－1－2)/(3+1)
3x+4y=5
CD直線方程式：4X－3Y=4*1－3*0.5=2.5
8x－6y=5
8x=6y+5
AB^2=4^2+3^2=25
AC^2=(p－3)^2+(q+1)^2=25
64(p－3)^2+64(q+1)^2=1600
(8p－24)^2+(8q+8)^2=1600
(6q+5－24)^2+(8q+8)^2=1600
36q^2－228q+361+64q^2+128q+64=1600
100q^2－100q－1175=0
4q^2－4q－47=0
q=(4+/-√(16+752))/8=(4+/-√768)/8=(4+/-16√3)/8=(1+/-4√3)/2
(1) q=(1+4√3)/2
8p=6q+5=3+12√3+5=8+12√3
p=(2+3√3)/2
C((2+3√3)/2，(1+4√3)/2)
(2) q=(1－4√3)/2
8p=6q+5=3－12√3+5=8－12√3
p=(2－3√3)/2
C((2－3√3)/2，(1－4√3)/2)