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Known vector A, B, and AB = A+2 B, BC=-5 A+6 B, CD =7 A-2 B, must be collinear () A.A, B, D B.A, B, C C.B, C, D D.A, C, D Known vector A, B, and AB = A+2 B, BC=-5 A+6 B, CD =7 A-2 B, then certain collinear () A.A, B, D B.A, B, C C.B, C, D D. A, C, D
∵
BD =
BC+
CD=-5
A+6
B+7
A-2
B=2
A+4
B=2
AB
∴
AB∥
BD
Because straight line AB and BD have common point B
So points A, B, D are in the same straight line.
Therefore, A.
Given the vector a=(1,2), b=(x,3) If a is perpendicular to b, find the value of x Seek, fast
From vectors a=(1,2), b=(x,3), a vertical b, x+6=0, so x=-6
From vectors a=(1,2), b=(x,3), a vertical b: x+6=0, so x=-6
The zero vector is parallel to any vector. Can you say that the zero vector is a parallel vector of any vector (the concept of parallel vector is non-zero)?
0
Let a b be two non-collinear non-zero vectors Why from the collinear of A, B and C, it can be known that there is real number λ, so that OC =λ OA +(1−λ) OB , Let a b be two non-collinear nonzero vectors Why from the collinearity of A, B and C, it can be known that there is real number λ, so that OC =λ OA +(1−λ) OB ,
The answer to the original question was not clear.
Since the vector OC =1/3(a+b),
If A, B and C are collinear, C must be on segment AB
So you can set:
OC =λOA+(1−λ) OB
Vector a-b = what (in coordinates) Vector a=(X1, Y1), vector b=(X2, Y2), vector a-b=:(expressed in coordinates)
0
Given point B (2,1), A (6,-3) and vector b=(1,3), find the angle between vector b and vector AB, and the projection of vector b on vector AB (By the way, how to solve such problems as angle and projection)
1 Vector AB =(2-6,1+3)=(-4,4) the angle θ between vector b and vector AB: cosθ= b • AB/ b AB =[1*(-4)+3*4]/[√(12+32)((-4)2+((-4)2+4 2)]=1/√5 so that θ is approximately equal to the projection of vector b on vector AB...
1 Vector AB=(2-6,1+3)=(-4,4) the angle θ between vector b and vector AB: cosθ=b • AB/ b AB =[1*(-4)+3*4]/[√(12+32)((-4)2+((-4)2+42)]=1/√5 so that θ is approximately equal to the projection of vector b on vector AB...
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