A few questions about vectors A question about vectors

1. Isn't the coordinate system all built......ob=(1，-1) oc=(1,1) od=(-1,1)

2. Let a=(x1,y1),b=(x2,y2)so x1+x2=2,x1-x2=-8,y1+y2=-8,y1-y2=16, respectively, to obtain a=(3,4) b=(5,-12).

3. Is the last vector e a vector c? If so, let c=ma+nb, so -2m+3n=10, 3m+n=-4, synthesis, the solution is m=-2, n=2, and the vector c=-2a+2b=2b-2a

，-1)oc=(1,1) od=(-1,1)

2.Let a=(x1,y1),b=(x2,y2)so x1+x2=2,x1-x2=-8,y1+y2=-8,y1-y2=16, respectively, to get a=(3,4) b=(5,-12).

3.Let c=ma+nb, so -2m+3n=10, 3m+n=-4, synthesis, the solution is m=-2, n=2, and the vector c=-2a+2b=2b-2a

1.It's very simple, the vertices of the square in the graph are symmetrical with respect to each coordinate axis, and since the starting point of each vector is the coordinate origin, ob=(1,-1) oc=(1,1) od=(-1,1) is obtained.

2.Two general equations can be given by adding two plane vectors with unknowns. From the title "vector a + vector b=(2,-8), vector a - vector b=(-8,16)", we can list the formula of adding two vectors, four general equations, from which we can find four unknowns, so we can set a=(x1,y1),b=(x2,y2) so x1+x2=2,x1-x2=-8,y1+y2=-8,y1-y2=16, respectively, to obtain a=(3,4) b=(5,-12).

3.First of all, we need to understand that any vector in the same plane can be represented by any two non-collinear vectors in that plane. So we can let c=ma+nb, so -2m+3n=10, 3m+n=-4, synthesis, solution m=-2, n=2, vector c=-2a+2b=2b-2a.

Vector BP = Vector BO + Vector OP

Vector PA = Vector PO + Vector OA

It is obtained from the vector bp = vector pa: vector bo + vector op = vector po + vector oa, that is, 2 times the vector op = vector oa + vector ob

Substituting the vector op=x by the vector oa+y by the vector ob yields: x=y=1 2

There are two solutions to this problem. First of all, draw the coordinate diagram according to the title, from the figure and the coordinates of each point, we can know that the angle between the vector A and B is 90 degrees, and the angle between B and the abscissa is 60 degrees, then the angle between C and the abscissa is 15 degrees, and because the modulus of C is the root number 2, so the coordinates of C are (root number 2cos15, root number 2sin15), that is, [root number 2 (root number 2) 4, root number 2-(root number 2) 4], and there is a solution in the third quadrant, but the symbols are negative.

This question has its peculiarities. Illustration: |a|=|b|.

It is easy to get a and b perpendicular. Therefore, isosceles right triangles can be constructed. Then the vector c must be the direction vector of the angular bisector of the right angle and the middle line of the bottom edge, that is, the vector is parallel to the straight line.

The midpoint can be found as ((root number 3 1) 2, (root number 3-1) 2) so c=k*((root number 3 1) 2,(root number 3-1) 2) and |c|= root number 2 so the solution k plus or minus 1 so c = plus or minus 1 * ((root number 3 1) 2, (root number 3-1) 2).

Let the vector c(m,n) have m*m +n*n=4; Then use the angle between vector A and vector C, and the angle between vector B and vector C is equal to list a set of equations. Finally, the vector c (2 roots, -2 roots) or (-2 roots, 2 roots) is obtained

Solution: (2a+b)*c=2a*c+b*c=(0,-3,-10)*(1,-2,-2)=26, and because a·c=4, b*c=26-8=18, the c vector obtains |c|=3，∴cos=（b*c）/|b||c|=1/2,∴=60°

Vector bc = (4, 3).

Let : ad=(Ming Hall x,y-2).

4=2x3=2(y-2)

So the sail track: x=2, y=

d(2,