The constant 0 is multiplied by a non-zero vector equal to the zero vector, then the zero vector is multiplied by a non-zero vector equal to zero or zero vector

The constant 0 is multiplied by a non-zero vector equal to the zero vector, then the zero vector is multiplied by a non-zero vector equal to zero or zero vector

The product of vectors is a number, so it is zero

The number of projectiles in vector a=(3,2) in vector b=(-3,4) is

Using Formulas
The projection is a*b/|b|
A*b=3*(-3)+2*4=-1
|B (9+16)=5
Projection =-1/5

In △ABC, if (vector AB)•(vector BC)+(vector AB)^2=0, the shape of △ABC can be determined as ()? Right triangle

Easy,
Extract vector AB
Vector AB (vector BC + vector AB)=0
So vector AB is multiplied by vector AC =0
AB vertical AC~!

In △ABC, the point M is the midpoint of BC, the point N is on the edge AC, and AN=2AC, AM and BN intersect at P, find the value of AP: PM. In △ABC, the point M is the midpoint of BC, the point N is on the edge AC, and AN=2AC, AM and BN are intersected by P, the value of AP: PM is obtained.

Is the question wrong? AN=2AC? AN=2NC bar PA=XAM, PB=YBN PC+CA=PAPC+CB=PBPA-CA=PB-CBXAM-CA=YBN-CBX (AB+BM)-CA=Y (BA+AN)-CBX (AC+CB+1/2BC)-CA=Y (BC+CA+2AC/3)-CBX (AC-1/2BC)+AC=Y (BC-AC/3)+BC (X+1) AC-X/2BC=-Y/3AC+(Y...

(Available on all letter groups →) Given that OA and OB are two non-collinear vectors, let the vector OM=λOA OB, and =1,λ, R Verification: M, A and B are collinear (Available on all letter sets →) Given that OA and OB are two non-collinear vectors, let the vector OM=λOA OB, and =1,λ, R Verification: M, A and B are collinear

OM=λOA OB=(1-μ) OA OB
OM-OA=μ(OB-OA)
AM=μAB
So:
M, A and B are collinear.

It is proved that if any non-zero vector in P^n is the eigenvector of n-order matrix A in number field P, then A must be a quantity matrix.

Ae1=a1e1, Ae2= a2e2,...,Aen =anen, where a1,a2,...,an is the eigenvalue,e1,e2,...,en is n columns of the unit matrix, so there is AE=ED, where D is the diagonal element of the a1,a2,...,an diagonal matrix.