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If the vector a=(1,-2),|b|=4|a|, and a, b are collinear, then the possible coordinates of b are If the vector a=(1,-2),|b|=4|a|, and a, b are collinear, then the possible coordinates of b are (-4,8)
B=±4a=±(4,-8)
When vector a and vector b satisfy what condition,|a+b|=|a-b|?
0
Why is it that vector a is multiplied by vector b directly by them and no module is multiplied by module to cos? The example is that the angle of unit vector e1, e2 is known to be 60°. Solve vector e1 point multiply e2=1/2 The next step is to multiply the vector a by the vector b=(2e1=e2)(2e2-3e1). I just don't understand here. Is n' t the vector a by the module of x vector b=a multiplied by the module of b multiplied by cos angle? How do you multiply the vectors a and b here? It's not the same definition ==
A·b=(2e1+e2)·(2e2-3e1)=-6|e1|^2+2|^2-2-4e1·e2
This is the distributive law of the product of the number of vectors
You can also use the formula: a·b=|2e1+e2 2e2-3e1|* cos <2e1+e2,2e2-3e1>
But this calculation is too cumbersome and flexible
A·b=(2e1+e2)·(2e2-3e1)=-6|e1|^2+2||e2|^2-2-4e1·e2
This is the distribution law of the product of the number of vectors
You can also use the formula: a·b=|2e1+e2 2e2-3e1|* cos <2e1+e2,2e2-3e1>
But this calculation is too cumbersome and flexible
Multiplication of vector moduli Given vector PA=(-6-X,-Y), vector PB=(6-X,-Y), help me simplify |PA| by |PB|=36. Multiplication of vector modules Given vector PA=(-6-X,-Y), vector PB=(6-X,-Y), help me simplify |PA| by |PB|=36.
| PA |=√[(x+6)^2+y^2]| PB |=√[(x-6)^2+y^2]
|PA PB|=√[(x^2+y^2+36)^2-144x^2]=36
X^4+y^4-72x^2+72y^2+2x^2y^2=0
The geometric meaning of vector product is a little unclear. How does |(a+2b)×(a-3b)| equal 5|a×b|? It's not very clear here.
Axb is the outer product of the vector, ab is the inner product of the vector, and the high school is the inner product
The outer product is a vector,|axb|=|a||b|sin
So |axa|=0,|bxb|=0, axb=-bxa
|(A+2b) x (a-3b)|=|axa-3axb+2bxa-6bxb|=|-3axb-2axb|=5|axb|
What is the geometric meaning of vector dot product?
In physics, the point product is used to represent the work done by the force. When there is an angle between the force F and the displacement S of the particle, the work done by the force F is W=—F—S—cos.
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