What are the problems that should be paid attention to in high school math vectors?

Attention should be paid to the direction indicated, and it is important to remember the algorithm.

The main problem is the projection, the magnitude and modulus of the vector, and then there is the parallelogram rule.

The addition and subtraction of vectors is equal to the addition and subtraction of the corresponding coordinates.

The multiplication of two vectors is equal to the multiplication and addition of the corresponding coordinates.

The condition for a b is x1*y2-x2y1=0a and the condition for b is x1*x2+y1*y2=0ab=(x2-x1,y2-y1).

The third question is collinear means that it is parallel, so you do it according to the sufficient and necessary conditions of parallelism, that is, it is.

ma=m*(1,2)=(m,2m)

ma+b=(m+2,2m+3)

ma+b) and c are collinear, that is, (m+2)*(7)-(4)*(2m+3)=0 solution. m=2

From the absolute value of the A vector * B vector = the absolute value of the A vector * the absolute value of the B vector, we know that cosa=1 is a = 0° (a is the angle between the vectors a and b), then the A vector and B vectors are in the same direction, A and B are in a straight line, and there are 2sin x = sin (2x).

2sin²x=2sinx cosx

Sin x = cos x or sinx=0 is known from the range x (0, ) of x: sinx=0 i.e. x=0° is not in line with the topic and should be discarded, then sin x = cos x i.e. x=45°, thus tan x = 1

There is a question that A and B are in a straight line.

So 2*(sin x) 2=sin (2x) gives sin x = cos x

Combined with the range of x to know.

x = 45°

tan x = 1

Passing C and A makes straight lines parallel to AB and BC, respectively. Let the intersection of two straight lines be d. Connect the OD.

Obviously, the vector od is the vector oa + vector oc. So the vector od=-3 vector ob. Thus o, b, d collinear.

Then make perpendicular lines from point C and A to OD respectively, which are denoted as AE and CFApparently AE and CF are equal in size (since the triangle AOD and the triangle COD area are equal). The areas of the triangle AOB and the triangle BOC are (OB times AE) 2 and (OB times CF) 2, respectively.

Thus the two areas are equal. The conclusion is 1

Too few points!

From Obiyama 3oa+4ob+5oC=0=>3oa+4ob=-5oC=> foolishness in OA*ob=0

oc*ab=oc(ob-oa)=-3 rot 5oa-4 5ob)(ob-oa)=-1 5.

From the vector ob + vector om = 3 * vector op-vector oa to obtain the vector ob-vector oa + vector om-vector oa = 3 * (vector op - vector oa).

That is, vector ab + vector am = 3 * to the nuclear oak ap

All P-points must be coplanar with ABM.

Condition 2 seems to be useless, the key is to go to the hole to drop the O point.

Vector ob + vector buried om = 3 * vector op - vector oa4 * vector oa - vector ob - vector om = vector op The addition of the two formulas ob = vector op ob = 4 * vector op

And because the point O is not on the flat buried surface ABM, the bent wide spike is not on the plane ABM.