It is proved that if vector a*b+b*c+c*a=0, then a, b, c are coplanar. What's the theorem, what's the definition, what's the idea

It is proved that if vector a*b+b*c+c*a=0, then a, b, c are coplanar. What's the theorem, what's the definition, what's the idea

Mainly the properties of outer product and mixed product operation:
A, b, c are coplanar if and only if (a, b, c)=0
(A, b, c)=(a×b)·c
(C, a, c)=0,
(B, c, c)=0
......
It is proved that if vector a×b+b×c+c×a=0,
Then (a×b+b×c+c×a)·c=0
(A, b, c)=0
So: a, b, c are coplanar

Let the column vector of matrix B be linearly independent, BA=C, and prove that the column vector of matrix C is linearly independent if and only if the column vector of matrix A is linearly independent.

On the one hand, it is obvious that the solution of AX=0 is the solution of CX=BAX=0. On the other hand, if X1 is the solution of CX=0, then CX1=0. So (BA) X1=0, so B (AX1)=0, because B column is full rank, there is AX1=0. That is, X1 is the solution of AX=0. So there is r (C)=r (A). Because A, C have the same number of columns, so C is...

Let a, b, c be three arbitrary vectors and prove that the vectors a-b, b-c, c-a are coplanar Math problem, help ___ Let a, b, c be three arbitrary vectors and prove that the vectors a-b, b-c, c-a are coplanar Math topic, help ___

Prove: Assume x, y, z with different coefficients of 0
Let (a-b)*x+(b-c)*y+(c-a)*z=0
I.e.(x-y)*a+(y-x)*b+(z-y)*c=0
When x = y = z is not equal to 0
(A-b)*x+(b-c)*y+(c-a)*z=0 holds
I. e. that vector a-b, b-c, c-a are coplanar

Proving that the sum of squares of four sides of a parallelogram is equal to the sum of squares of two diagonals Solution by the formula of distance between two points

This is also proved by vectors and complex numbers.

Prove that the shape of parallelogram remains unchanged during translation and rotation, so only one special case is proved: A (x1, y1), B (x1+a, y1).C (x1+a+b, y1+h), D (x1+b, y1+h), as shown in the figure:

It is proved by the vector method that the sum of the bisectors of the two diagonals of a parallelogram is equal to twice the sum of the squares of the two adjacent sides It is better to have a process One hundred thousand. Thank you! The first floor... Is the vector method that simple? ==||| Feel like the process is so short... No... Of course I think it's better to keep it short... But the teacher won't let... But forget it... Thank you for your quick answer... ==||| I hate my winter homework... Ahhh...

AB + AC = AD AB - AC = BC square plus
Of course. Not so short.

Urgent! Proving that the sum of squares of diagonals of parallelograms is equal to the sum of squares of four sides RT I would like to explain it in more detail. If it involves some trigonometric functions, I also want to explain it.

Method 1: Make the height of BC edge through two points A and D. If the perpendicular foot is E and F respectively, then △ADE △DCFBE=CF, AE=DF, using Pythagorean theorem, we can get square=BF square+DF square BD square=(BC+CF) square+DF square=BC square+2*BC*CF+CF square+DF square AC square=AE square+CE square=AE square+(BC-BE) square=...