中五附加數學----Vectors in the Two-Dimensional Space

1.Find (→a)‧(→b) in each of the following.
(a)(→a)=2(→i) - 3(→j),b = 5(→i) - 2(→j)
(b)a = -4(→i)+(→j),b = -2(→i) - 2(→j)
(a)
(→a)‧(→b)=2*5+(-3)(-2)=16
(b)
(→a)‧(→b)=(-4)(-2)+(-2)=6
2
c^2=a^2+b^2-2abcosC
cosC=(a^2+b^2-c^2)/2ab
cosC=(36+25-16)/(2*6*5)=0.75
(→a)‧(→b)
=|a||b|cosC
=6*5*0.75
=22.5
3
(a)
(→DC)=(→AB)=a
(→BC)=(→AD)=b
(→AB)+(→BD) =(→AD)
a+(→BD)=b
(→BD) =b-a
(→AC)
= (→AB) + (→BC)
=a+b
(b)
(AC)*+(BD)*
=(→AC) ‧(→AC) +(→BD)‧(→BD)
=(a+b)(a+b)+(b-a)(b-a)
=(|a|*+ |b|* +2ab)+(|b|*+ |a|* -2ab)
=2( |a|*+ |b|* )

Remember that if a=xi+yj,b=mi+nj
Then a．b=mx+ny
(I won't put the symbol →)
(1)
(a)
a=2i-3j, b=5i-2j
Therefore a．b=(2)(5)+(-3)(-2)=10+6=16

(b)
a=-4i+j, b=-2i-2j
Therefore a．b=(4)(-2)+(1)(-2)=-8-2=-10

(2)
Because the vector sum of the vectors is zero.
Therefore the vectors make a triangle.
Let the angle between a and b be θ
|c|^2=|a|^2+|b|^2-2|a||b|cos θ
6^2+5^2-2(6)(5)cos θ=4^2
61-60 cos θ=16
60 cos θ=45
cos θ=3/4

(3)
(a)
Because AB=AC,AD=BD
Therefore BD=b and AC=a

(b)
I do not understand what is the meaning of (*=2), I am sorry for that.