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Given that the angle between vector a and vector b is 120 degrees and the module of vector a = the module of vector b =4, then the value of vector b multiplied (2 multiplied by vector a + vector b) is:
First you understand the vector multiplication formula: a · b =| a || b | cos α(α is the angle between them)
The title then knows that α=120°;| a|=| b|=4
So
B·(2·a+b)
=2·A·b+b·b
=2| A | b |· cos120 b |^2 cos0°
=2*4*4*(-1/2)+4*4*1
=-16+16
=0
Familiarity with formulas is key!
↖(^ω^)↗
Vector a=(2,1), b=(3, x), if (2a-b) is perpendicular to b, then x is? Just think and answer.
The dot product of vector (2a-b) and vector b is zero, and the coordinate is substituted and the answer is x=-1 or 3
If the dot product of vector (2a-b) and vector b is zero, substitute the coordinates into and the answer is x=-1 or 3
Given vector a=(cosx, sinx), vector b=(√3,-1), then |2a vector-b vector | has the maximum value and the minimum values are? Given vector a=(cosx, sinx), vector b=(√3,-1), then |2a vector-b vector | has the maximum value and the minimum values are?
|Vector 2a-vector b|2=(2cosx-√3)2+(2sinx+1)2=4cos2x+4sin2x+3+1-4√3cosx+4sinx=8+4(sinx-√3cosx)=8+8sin (x-π/3),|vector 2a-vector b|=under the root sign (8+8sin (x-π/3)),8sin (x-π/3)∈[...
Known vector A =(cosθ, sinθ), vector B =( 3,−1), Then |2 A− B|, the minimum values are ___.
2
A-
B=(2 cosθ-
3,2Sin 1),|2
A-
B |=
(2 Cosθ-
3)2+(2Sin 1)2=
8+4Sinθ-4
3 Cosθ=
8+8S in (θ-π
3),
4 Max,0 min
Therefore, the answer is:4,0.
Given vector a=(cosa, sina), vector b=(root 3,-1), then the maximum value of |2a-b|? Detailed process
|A|=1,|b|=2
|2A-b|2=4a2-4ab+b2
=4-4(√3 Cosa-sina)+4
=-8(√3/2Cosa-1/2sina)+8
=-8Cos (a+30°)+8≤8+8=16
The maximum value of 2a-b| is √16=4
Given vector a=(cosF, sinF), vector b=(root number 3,-1), then the maximum value of vector | of vector -b of |2a
2A vector ={2cosF,2sinF} vector 2a-b ={2cosF-√3,2sinF+1}|2a vector -b vector |=√(2cosF-√3)^2+(2sinF+1)^2=√[4(cosF)^2-4√3cosF+3+4(sinF)^2+4sinF+1]=√{4[(sinF)^2+(cosF)^2]+4sinF -4√3cos...
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