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Mathematics vector problem in senior one: solution and answer
1, this topic examines the vector * * * straight line, the necessary and sufficient conditions of equality, and the basic theorem of plane vector.

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).