Given vectors OA, OB, OC and trilateral a, b, c, I are the inner of a triangle, prove vector OI=(aOA+bOB+cOC)/(a+b+c) Given vectors OA, OB, OC and trilateral a, b, c, I are the inner of a triangle, prove that vector OI=(aOA+bOB+cOC)/(a+b+c)

Given vectors OA, OB, OC and trilateral a, b, c, I are the inner of a triangle, prove vector OI=(aOA+bOB+cOC)/(a+b+c) Given vectors OA, OB, OC and trilateral a, b, c, I are the inner of a triangle, prove that vector OI=(aOA+bOB+cOC)/(a+b+c)

Let the triangle ABC, AD be the angle bisector on the edge of BC, the inner is I.|BC|=a,|AC|=b,|AB|=caIA+bIB+cIC=aIA+b (AB+IA)+c (AC+IA)=(a+b+c) IA+b (DB-DA)+c (DC-DA) Let the direction vector e of BC, then DB=e|DB|, DC=-e|DC| is the angle bisector theorem.

Point O is the center of the inscribed circle of △ABC, abc is the length of the opposite side of ∠A∠B∠C, and aOA+bOB+cOC=0(OAOBOC and 0 are vectors) The point O is the center of the inscribed circle of △ABC, and abc is the length of the opposite side of ∠A∠B∠C Verify aOA+bOB+cOC=0(OAOBOC and 0 are vectors)

CERTIFICATE:
A=OB-OC
B=OC-OA
C=OA-OB
Then a*OA+b*OB+c*OC=(OB-OC)*OA+(OC-OA)*OB+(OA-OB)*OC
Expand to get proof!
Explain that all the above are vectors,* is point multiplication not X multiplication

Given that O is a point in triangle abc, AB=c, BC=a, AC=b, if aOA+bOB+cOC=zero vector,(OA, OB, OC all vectors) prove that O is inner.

All we use are vectors, where e is the unit vector,
Let AB=c*e1, AC=b*e2, BC=a*e3
Where e1, e2, e3 are unit vectors in the ABACBC direction,
AOA+bOB+cOC=0
I.e. aOA+b (OA+AB)+c (OA+AC)=0
I.e.(a+b+c) OA+bAB+cAC=0
So OA =-(bAB+cAC)/(a+b+c)=-(bc*e1+bc*e2)/(a+b+c)=-bc (e1+e2)/(a+b+c)
I.e. OA=u (e1+e2), where u=-bc/(a+b+c)
So OA is the angular bisector of the angular BAC.
Similarly, it can be proved that OB is the angle bisector of angle ABC, OC is the angle bisector of angle ACB,
So O is the heart of the triangle.

3 Let OA, OB be a non-collinear vector, if OP=aOA+bOB (a, b∈R), find the necessary and sufficient conditions for three points A, B, P to be collinear OAOBOP is vector

PA=OA-OP=(1-a) OA-bOB
PB=OB-OP=(1-b) OB-aOA
Three points A, B, P collinear
PA=nPB
(1-A) OA-bOB=n [(1-b) OB-aOA]
-B/(1-b)=(1-a)/(-a)
(1-A)(1-b)=ab
1-A-b=0
A+b=1

Conditions for Three Vectors to Form a Triangle The for Three Vectors to Form a Triangle

The three vectors are not parallel in pairs and their sum is 0.

Why are three non-zero vectors coplanar if and only if the determinant of these three vectors equals zero

It's easy to think. Three vector determinants are zero, which means that the matrix composed of three vectors is not full-rank, that is to say, in the maximal independent group of vector groups, the number of vectors is less than 3, that is, there must be vectors that can be expressed linearly by other vectors, which is not to say that three vectors are coplanar.

It's easy to think. Three vector determinants are zero, which means that the matrix composed of three vectors is not full-rank, that is to say, in the maximal independent group of vector groups, the number of vectors is less than 3, that is, there must be vectors that can be expressed linearly by other vectors.