A, b non-zero vector, satisfy |a+b|=|a-b| Why is the vector a vertical vector b?

A, b non-zero vector, satisfy |a+b|=|a-b| Why is the vector a vertical vector b?

Basic concept:|a+b|=|a-b|, i.e.:|a+b|^2=|a-b|^2, and:|a+b|^2=(a+b)·(a+b)=|a 2 b|^2+2a·b||a-b|^2=(a-b)·(a-b)=|a 2 b|^2-2a·b, so:|a 2 b|^2+2a·b=|a 2 b|^2-2a·b is: a·b=0, so: a and b are vertical...

| Vector a|=0 vector a = zero vector

See definition: zero vector definition: there are two definitions of "vector with modulus length of zero ":1) is vector,√2) is modulus length of zero? √ Both of them meet the definition of zero vector, so "|a|=0, then a=0. represents vector." is the true proposition!

One had to look at the definition: Zero vector definition:" Vector with modulus length of zero "There are two definitions:1) is a vector,√2) is modulus length of zero? √ Both of them met the definition of zero vector, so "|a|=0, then a =0. Representing the vector." This was the real question!

Given vector a=(2,-1,3), b (-1,4,-2), C=(7,5, in), if three vectors a, b, c are coplanar, then the real number in is equal to

Because the three vectors a, b, c are coplanar
So
2X-y=7
-1X+4y=5
λ=3X-2y
Jiede
X =33/7
Y =17/7
λ=65/7

Given that the coordinates of point A are (-2,1) point B as (3,5), find the coordinates of vector AB and vector BA

Vector AB=(3+2,-5-1)=(5,-6)
Vector BA=- Vector A=(-5,6)

Given that three points A (4,-2), B (-4,4), C (1,1), C as vector CD and vector AB as collinear, and the module of vector CD=5, then the coordinate of point D is Given that three points A (4,-2), B (-4,4), C (1,1), and the point C passing through of vector CD and vector AB, and the module of vector CD=5, then the coordinate of point D is

Set point D (x, y)
(Y-1)/(x-1)=(-2-4)/(4+4)=-3/4
Y=-3x/4+7/4
(X-1)2+(-3x/4+7/4-1)2=25
(X-1)2=16
X-1=±4
X=5 y=-2
Or x=-3 y=4
So point D is (5,-2) or (-3,4)

P={α/α=(-1,1)+M (1,2) xy∈R} Q={β/β=(1,-2)+n (2,3) n∈R} is two sets of two vectors, then P∩Q?

Joint PQ
(-1,1)+M (1,2)=(1,-2)+n (2,3)
(-1+M,1+2M)=(1+2n,3n-2)
M-1=1+2n
2M+1=3n-2
Jiede
N=-7
M=-12
Take back P or Q, get (-13,-23)
P∩Q=(-13,-23)

Joint PQ
(-1,1)+M (1,2)=(1,-2)+n (2,3)
(-1+M,1+2M)=(1+2n,3n-2)
M-1=1+2n
2M+1=3n-2
Solved
N=-7
M=-12
Take back P or Q, get (-13,-23)
P∩Q=(-13,-23)