In triangle ABC, O is a point in plane, and let vector OA=vector a, vector OB=vector b, vector OC=vector c Then satisfy the condition (vector a + vector b)• vector AB=(vector b + vector c)• vector BC=(vector c + vector a)• what is the center of the triangle when vector CA In triangle ABC, O is a point in plane, and let vector OA=vector a, vector OB=vector b, vector OC=vector c If the condition (vector a + vector b)• vector AB =(vector b + vector c)• vector BC =(vector c + vector a)• vector CA, what center is O of the triangle In triangle ABC, O is a point in plane, and let vector OA=vector a, vector OB=vector b, vector OC=vector c If the condition (vector a + vector b)• vector AB=(vector b + vector c)• vector BC=(vector c + vector a)• vector CA, what is the center of the triangle

In triangle ABC, O is a point in plane, and let vector OA=vector a, vector OB=vector b, vector OC=vector c Then satisfy the condition (vector a + vector b)• vector AB=(vector b + vector c)• vector BC=(vector c + vector a)• what is the center of the triangle when vector CA In triangle ABC, O is a point in plane, and let vector OA=vector a, vector OB=vector b, vector OC=vector c If the condition (vector a + vector b)• vector AB =(vector b + vector c)• vector BC =(vector c + vector a)• vector CA, what center is O of the triangle In triangle ABC, O is a point in plane, and let vector OA=vector a, vector OB=vector b, vector OC=vector c If the condition (vector a + vector b)• vector AB=(vector b + vector c)• vector BC=(vector c + vector a)• vector CA, what is the center of the triangle

(Vector a + vector b)• vector AB =(vector b + vector c)• vector BC =(vector c + vector a)• vector CA,
——》(Vector a + vector b)•(vector b - vector a)=(vector b + vector c)•(vector c - vector b)=(vector c + vector a)•(vector a - vector c)
——》 Vector b• vector b - vector a• vector a = vector c• vector c - vector b• vector b= vector a• vector a - vector c• vector c,
——》 Vector a• vector a= vector b• vector b= vector c• vector c,
——》 Vector a = vector b = vector c,
——》 O is the vertical center (or outer center) of the triangle.

If O is a point in the plane of triangle ABC and satisfies vector OA+2 vector OB+3 vector OC=0, then the area ratio of triangle AOC to triangle BOC is 2:1, why? If O is a point in the plane of triangle ABC and satisfies the vector OA+2 vector OB+3 vector OC=0, then the area ratio of triangle AOC to triangle BOC is 2:1, why?

Extend OB to B' to OB'=2OB; extend OC to C' to OC'=3OC;
Link B'C', take point D in B'C', link OD and extend to A'so that DA'= OD;
Connecting B'A', C'A', the quadrilateral OB'A' C' is a parallelogram
2 Vector OB +3 vector OC = vector OB'+ vector OC'= vector OA'
Further vector OA+2 vector OB+3 vector OC=0
I.e. vector OA+vector OA'=0, vector AO=vector OA'
So A, O, A' are collinear, and |AO|=|OA'|
Using a triangle of the same height as the base, equal in area:
S△AOC = S△A'OC = S△OCB'=2S△BOC ==> S△AOC / S△BOC =2/1

Given that there are four points OABC on the plane satisfying vector OA+OB+OC=0 vector, vector OA*OB=OB*OC=OC*OA=-1, what is the perimeter of triangle ABC? The answer is 9 Why?

Vector OA+OB+OC=0,
O is the center of gravity of △ABC,
Vector OA*OB=OB*OC=OC*OA=-1,
OA*BC=OA*(OC-OB)=OA*OC-OA*OB=0,
OA⊥BC,
Similarly, OB⊥CA,
O is the vertical center of △ABC,
ABC is an equilateral triangle,
∠BOC=120°,
OB*OC=|^2cos120°=(-1/2)|OB|^2=-1,
,|OB|^2=2,|OB||=√2,|BC||=|OB||√3=√6,
Perimeter of triangle ABC =3√6.
You gave the wrong answer.

It is known that ABC is a non-collinear three point in the plane, o is the center of gravity of △ABC, and the moving point p satisfies the vector OP=1/3(1/2 vector OA+1/2 vector OB+1/2 vector OC), Then point P must be △ABC It is known that ABC is a three point which is not collinear in plane, o is the center of gravity of △ABC, and the moving point p satisfies the vector OP=1/3(1/2 vector OA+1/2 vector OB+1/2 vector OC), Then point P must be △ABC It is known that ABC is a three point which is not collinear in plane, o is the center of gravity of △ABC, the moving point p satisfies vector OP=1/3(1/2 vector OA+1/2 vector OB+1/2 vector OC), Then point P must be △ABC

P must be the center of gravity of the triangular ABC.
This is because O is the center of gravity, then OA+OB+OC=0 vector, so OP=0 vector, so P coincides with O.

P must be the center of gravity of the triangle ABC.
This is because O is the center of gravity, then OA+OB+OC=0 vector, so OP=0 vector, so P coincides with O.

Given that the three points A, B and C are not collinear, for the point O outside the plane ABC, if the vector OP=2 vector OA-vector OB-vector OC, whether the point P is coplanar with A, B and C

The necessary and sufficient condition of coplanarity is that for any point O in the space, there is a vector OP=m•OA+n•OB+s•OC, where m+m+s=1 Because OP=2OA-OB-OC,2-1-1=0=1 in this question, there is no coplanarity.

The necessary and sufficient condition for the coplanarity of P, A, B, C is that for any point O in a space, there is a vector OP=m•OA+n•OB+s•OC, where m+m+s=1. In this case, OP=2OA-OB-OC,2-1-1=0=1, so that it is not coplanar.

The necessary and sufficient condition of coplanarity is that for any point O in the space, there is a vector OP=m•OA+n•OB+s•OC, where m+m+s=1 In this case, OP=2OA-OB-OC,2-1-1=0=1, so that it is not coplanar.

Given that the three points A, B and C are not collinear, for any point O outside the plane ABC, the conditional vector OP=1/5 vector OA+2/5 vector OB+2/5 vector OC is satisfied, Try to judge whether P is coplanar with A, B and C

Remember the conclusion
OP=xOA+yOB+zOC
If and only if x+y+z=1 for PABC coplanar
1/5+2/5+2/5=1
P is coplanar with A, B, C

Remember the conclusion
OP=xOA+yOB+zOC
PABC is coplanar if and only if x+y+z=1
1/5+2/5+2/5=1
P is coplanar with A, B, C