Solution: Suppose d= zero vector, then a=-kb, then lines A and b*** contradict the known ones.
Therefore, the vector d is nonzero. From the necessary condition of vector line, it is concluded that there is a real number m, so that C = MD
That is ka+b=m(a+kb)=ma+mkb,
That is, (k-m) vector a+( 1-mk) vector b=0 vector.
According to the basic theorem of plane vector, the above formula holds if and only if k-m= 1-mk=0.
Solution: k 2 = 1
That is k=+ 1 or-1.
2. This topic examines the definition of vector quantity product and the solution of vector modulus.
The product of c and b is c*d=(2a-b)*(3b-a).
=6a*b-2a^2-3b^2+a*b
= 7 *1*1* cos120 degrees -2 * 1 2-3 * 1 2
=(-7/2)-5
=- 17/2
Because C2 = (2a-b) 2 = 4a2-4a * b+B2 = 5-4 * (1/2) = 7.
So the modulus of c = root number 7.
Similarly, the modulus of b = root sign 13.
So the modulus of cos < c, d > = the modulus of c * d/c * d.
=(- 17/2)/ radical number 9 1
=- 17 root number 91182
3,
This topic examines the coordinate operation of vectors.
Proof (1): let A = (A 1, A2) and B = (B 1, B2), then Ma+NB = (Ma 1+NB 1, MA2+NB2).
Known slave, left = (ma2+nb2,2 [ma2+nb2]-[ma1+nb1])
Right = m (a2 a2,2a2-a1)+n (B2,2b2-b1)
=(ma2+na2，2[ma2+nb2]-[ma 1+nb 1])
So left = right.
That is, the original formula holds.
Proof (2): f (a) = (1, 2- 1) = (1, 1)
f(b)=(0，0- 1)=(0，- 1)
Let the coordinate of vector c be c(x, y), then
f(c)=(y，2y-x)=(p，q)
That is, y = p.
2y-x=q
Solution: x=2p-q
y=p
So the coordinate of vector c is (2p-q, p).