The quantity product problem of vector: if a=3i+2j-k, b=i-j+2k, then the quantity product of 5a and 3b is equal to The quantity product is not |a||b|Why is it directly 5a*3b=45-30-30=-15 Do not multiply by angle! The quantity product of vectors: if a=3i+2j-k, b=i-j+2k, then the quantity products of 5a and 3b are equal to The quantity product is not |a||b|Why is it directly 5a*3b=45-30-30=-15 Do not multiply by angle!

The quantity product problem of vector: if a=3i+2j-k, b=i-j+2k, then the quantity product of 5a and 3b is equal to The quantity product is not |a||b|Why is it directly 5a*3b=45-30-30=-15 Do not multiply by angle! The quantity product of vectors: if a=3i+2j-k, b=i-j+2k, then the quantity products of 5a and 3b are equal to The quantity product is not |a||b|Why is it directly 5a*3b=45-30-30=-15 Do not multiply by angle!

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A, B coordinates, how to find the vector AB How to find the vector AB How to find the vector AB?

Given A (a, b), B (c, d) then vector AB=(c-a, d-b)

Given two vector coordinates (a, b) and (c, d), ask how their multiplication is represented

Multiply by ac + bd
Cross-multiplication multiplies their modular product by a two-vector angle sine

Multiply by ac + bd
Cross multiplication is the product of their modules multiplied by a two-vector angle sine

The coordinates of the space vector operation represent the coordinates of points D that satisfy the following requirements, given A (1,0,0), B (0,0,1), and C (0,0,2 DB⊥AC, DC⊥AB and AD=BC

The basic idea is:
Let D (x, y, z);
Then vector BD (x, y, z-1);
Vector CA (1,0,-2);
Vector CD (x, y, z-2);
Vector BA (1,0,-1);
Vector AD (x-1, y, z);
Vector BC (0,0,1);
Then vector product (BD, CA)=0
Vector product (CD, BA)=0
(X-1)*(x-1)+y*y+z*z=1;
Three equations, find x, y, z.
However, I don't know if there is a problem with the topic of LZ, or if there is a flaw in my process. I ca n' t solve it. Please check it with LZ.

The Standard Orthogonal Decomposition and Coordinate Representation of Space Vector Suppose that the unit vectors e1, e2, e3 are perpendicular to each other, and the orthogonal decomposition of the vector a in the direction of e1, e2, e3 is 3e3-2e3-4e2, and the projection of the vector a in the direction of e1, e2, e3 is obtained respectively The Standard Orthogonal Decomposition and Coordinate Representation of Space Vector Let the unit vectors e1, e2, e3 be perpendicular to each other, and the orthogonal decomposition of the vector a in the direction of e1, e2, e3 be 3e3-2e3-4e2, so as to obtain the projection of the vector a in the direction of e1, e2, e3, respectively

E1=0
E2=4
E3=1
Almost here.
You guys are learning fast!

E1=0
E2=4
E3=1
It's almost here.
You guys are really fast!

A matrix of rank 1 that must be decomposed into the form of a column matrix (vector) and a row matrix (vector A matrix of rank 1 that must be decomposed into the form of a column matrix (vector) and a row matrix (vector R (A)=1, so it is convenient to set A = T and then calculate A^n in this way... A matrix of rank 1 that must be decomposed into the form of a column matrix (vector) and a row matrix (vector Why is that? A matrix of rank 1 that must be decomposed into the form of a column matrix (vector) and a row matrix (vector A matrix of rank 1 that must be decomposed into the form of a column matrix (vector) and a row matrix (vector R (A)=1, so it is convenient to set A = T and then calculate A^n... A matrix of rank 1 that must be decomposed into the form of a column matrix (vector) and a row matrix (vector Why is that? A matrix of rank 1 that must be resolved into the form of a column matrix (vector) and a row matrix (vector A matrix of rank 1 that must be resolved into the form of a column matrix (vector) and a row matrix (vector R (A)=1, so it is convenient to set A = T and then calculate A^n... A matrix of rank 1 that must be resolved into the form of a column matrix (vector) and a row matrix (vector Why is that?

CERTIFICATE:
If the rank of A is 1, let's suppose that the kth column of A is nonzero and is denoted by α.
Then all other columns of A can be expressed by α linear table, i.e. existence number
B1,b2,b3,...,bn makes
A1=b1α, a2=b2α,..., an=bnα,
Where a1,a2,...,an is column 1,2,..., n of A.
Let β=(b1,b2,...,bn)^T, then
A =(a1,., an)
=(B1α, b2α,..., bnα)
=α(B1,b2,...,bn)
=αβ^ T