If O is a point in the plane of triangle ABC and iob-oci = iob + oc-2oai is satisfied, then the shape of triangle ABC is All the letters in the question represent vectors

If O is a point in the plane of triangle ABC and iob-oci = iob + oc-2oai is satisfied, then the shape of triangle ABC is All the letters in the question represent vectors

Let o be the origin. OA, OB and OC are represented by a, B and C respectively. Of course, they are all vectors
Then there is
｜b-c|=|b+c-2a|
Square has, tidy up
a^2-(a,c)-(a,b)+(b,c)=0
That is (A-C, a-b) = 0
That is to say, vectors a-c,. A-B are perpendicular to each other, that is, CA is perpendicular to ba,
This triangle is a right triangle

In the triangle ABC, O is a moving point of the midline am. If am = 2, what is the minimum value of the vector OA (OB + OC)?

O is a moving point of the midline am,
According to the parallelogram rule, OB + OC = 2m,
OA•(OB+OC)= OA•2OM=2|OA||OM|cos180°
=-2|OA||OM|
According to the basic inequality, it can be concluded that:
|OA||OM|≤((|OA|+|OM|)/2)²=（|AM|/2）²=1,
-2|OA||OM|≥-2,
The minimum value of vector OA (OB + OC) is - 2

If the nonzero vectors OA and ob are not collinear, and the vector 2op = xoa + yob, if the vector PA = cab and C are real numbers, then what is the trajectory equation of point Q (x, y)

Vector PA = oa-op = OA - (xoa + yob) / 2 = (1-x / 2) OA - (Y / 2) ob, vector AB = ob-oa, from vector PA = cab, (1-x / 2) OA - (Y / 2) ob = - COOA + COOB, non-zero vector OA, ob is not collinear, ■ {1-x / 2 = - C, - Y / 2 = C}, eliminate C to get 1-x / 2 = Y / 2, simplify to get y = - x + 2, as the solution