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Plane vector. Given (a+b) vertical (2a-b),(a-2b) vertical (2a+b), try to find the cosine value of the included angle of a, b. Plane vector Given (a+b) vertical (2a-b),(a-2b) vertical (2a+b), try to find the cosine of the included angle of a, b. "Trouble detail" Thank you so much! Plane vector. Given (a+b) vertical (2a-b),(a-2b) vertical (2a+b), try to find the cosine value of the included angle between a and b. Plane vector. Given (a+b) vertical (2a-b),(a-2b) vertical (2a+b), try to find the cosine value of the included angle between a and b. "Please be more detailed." She was endlessly grateful!
0=(A+b)(2a-b)=2|a|^2+ab-||b|^2,0=(a-2b)(2a+b)=2|a|^2-3ab-2||b|^2,3[||b|^2-2||a|^2]=3ab=2||a|^2-2||b|^2,||b|^2=(8/5)|^2, ab=2/5||a|^2, cosine=ab/[|a||b||]=(2/5)/(8/5)^(1/2)=-(10)^(-1/2)
Vector A and vector B are known to be non-zero vectors, and A+B is perpendicular to 2A-B, and 2A-B is perpendicular to A+2B. ( The arrow on my vector can't hit, or can I read it? The problem solving process is most important. Vector A and vector B are known to be non-zero. And A+B is perpendicular to 2A-B and 2A-B is perpendicular to A+2B. ( The arrow on my vector can't be hit, or can I read it? The problem solving process is most important.
A+B perpendicular to 2A-B
(A+B)(2A-B)=0 i.e.2A^2+AB-B^2=0
2A-B vertical to A+2B
(2A-B)(A+2B)=0 i.e.2A^2+3AB-2B^2=0
Subtract 2AB = B^2
Two-vector collinear
So angle cosine is 1
A+B perpendicular to 2A-B
(A+B)(2A-B)=0 i.e.2A^2+AB-B^2=0
2A-B vertical to A+2B
(2A-B)(A+2B)=0 i.e.2A^2+3AB-2B^2=0
2AB = B^2
Two-vector collinear
So angle cosine is 1
Simplify 5(2a-2b)+4(2b-3a) Note that ab is a vector!
Strange, but the answer is 3a-2b.
Given that a, b are non-collinear vectors, and vector AB=3a+2b, vector CB=a+λb, vector CD=-2a+b, if A, B, D are collinear, try to find the value of real number λ
BD=CD-CB=-2a+b-a-λb=-3a+(1-λ) b
λ=3 Is obtained from A, B, D collinear 2/(1-λ)=3/(-3)=-1
BD=CD-CB=-2a+b-a-λb=-3a+(1-λ) b
λ=3 From A, B, D collinear 2/(1-λ)=3/(-3)=-1
BD=CD-CB=-2a+b-a-λb=-3a+(1-λ) b
λ=3 Is obtained from A, B, D collinearly 2/(1-λ)=3/(-3)=-1
Given that vector a is a unit vector starting from point A (3,-1) and perpendicular to vector b=(-3,4), then the end point coordinate of vector a is?
A (3, y)(y=-1)
Given vector a=(3,4), vector b is perpendicular to vector a, and vector b starts from (1,2) and ends from (x,3x), then vector b equals
(3,4)(X-1,3x-2)=0, i.e.3x-3+12x-8=0, x=11/15, b=(-4/15,3/15)
RELATED INFORMATIONS
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- 15. Given vector m=(sinA, cosA), n=(root number 3,-1), m*n=1, and A is an acute angle, find the size of angle A
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- 17. Given a=(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of real number m is [] Given a=(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of real number m is [] A and b are vectors (A)1(B)-1 (C)±1(D) without solution Given a =(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of the real number m is [] Given a =(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of the real number m is [] A and b are vectors (A)1(B)-1 (C)±1(D) without solution
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