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Is any non-zero vector parallel to the unit vector
Any non-zero vector is composed of two parts: modulus length and direction
The unit vector only has a modulus length of 1, but the direction can be arbitrary
So it's not necessarily parallel.
Any non-zero vector is composed of two parts: modulus length and direction
The unit vector is only 1, but the direction can be any direction
So it's not necessarily parallel.
Let vector a and vector b be a set of substrates in the plane, and prove that: When λ1a 2b=0,λ1=λ2=0.
Because vector a, vector b is a set of substrates in the plane.
So their moduli are equal and the two vectors are not collinear...
And because λ1a 2b=0... if a, b are collinear... then λ1 2=0
But because a, b are not collinear...
Then constant λ1=λ2=0...
Given that a and b are a set of bases representing all vectors in a plane, why can't a vector a and (a+b) vectors act as a set of bases? Given that a and b are a set of bases representing all vectors in a plane, why are the a and (a+b) vectors not a set of bases?
Let R (a, b)=m
Define matrix L =[1,0;1,1]
Then (a, a+b)= L (a, b)
Because L is reversible, R (a, a+b)=m
Therefore (a, a+b) can be used as the base vector group
The vectors e1, e2 are known as a set of bases of all vectors in the plane a,(as follows) And a=e1+e2, b=3e1-2e2, c=2e1+3e2, if c=λa b,(λ, R), try to find the values of λ,μ. I've done it. Maybe it's wrong. It's not the answer. It's clear.
C=λa b=λ(e1+e2)(3e1-2e2)=(λ+3μ) e1+(λ-2μ) e2=2e1+3e2
So λ=13/5,μ=-1/5
In the following group of vectors, the () A. A =(0,0), B=(1,-2) B. A=(-1,2), B=(5,7) C. A =(3,2), B=(6,4) D. A =(2,8), B =(1,4)
A: zero vector is collinear with any vector, so a=(0,0), b=(1,-2) can not represent the basis of all vectors in their plane; B:-1×7-2×5=-17=0, a=(−1,2), b=(5,7) can represent the basis of all vectors in their plane; C:3×4-2×6=0, a=(3,2...
Let e1, e2 be a set of bases of all vectors in a plane, then (), which can not be used as bases of the following four sets of vectors, is. A.e1+e2 and e1-e2 B.3e1-2e2 and 4e2-6e1 C.e1+2e2 and e2+2e1 D.e2 and e1+e2
B.3e1-2e2 and 4e2-6e1
-2(3,-2)=(-6,4)
That is, the two are parallel.
So it can not be used as a substrate.
B.3e1-2e2 and 4e2-6e1
-2(3,-2)=(-6,4)
That is, the two are parallel.
So it can't be the base.
RELATED INFORMATIONS
- 1. If e1 and e2 are a set of substrates in the plane, the following four sets of vectors can be used as the substrates of the plane vector: A, e1-e2, e2-e1B,2e1-e2, e1-1/2e2C,2e2-3e1,6e1-4e2D, e1+e2, e1-e2 If e1, e2 are a set of bases in the plane, the following four sets of vectors can be used as the bases of the plane vector: A, e1-e2, e2-e1B,2e1-e2, e1-1/2e2C,2e2-3e1,6e1-4e2D, e1+e2, e1-e2
- 2. 0
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