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Known vector A =(1,2), B=(-2,3), C=(4,1), if used A and B indicates C, then C =___.
Set
C=m
A+n
B
Then (4,1)=(m,2m)+(-2n,3n)
M−2n=4
2M+3n=1
N=-1, m=2
Therefore, the answer is:
C=2
A−
B
(Math) Given a=(10,-4), b=(3,1), c=(-2,3), try base b, c=a (letters represent vectors) Urgent! !! Fast return (Math) Given a=(10,-4), b=(3,1), c=(-2,3), try base b, c=a (letters represent vectors) Urgent! !! Quick return
3M-2n=10
M+3n=-4
Solve m=22 to 7
N=-2 to 7
A=7/22b-7/2c vector
3M-2n=10
M+3n=-4
Solution: m=22 to 7
N=-2 to 7
A=7/22b-7/2c vector
Given vector a=(2,-4), vector b=(-1,3), vector c=(6,5), vector p=vector a+2 times vector b-vector c, based on vectors a, b Given vector a =(2,-4), vector b =(-1,3), vector c=(6,5), vector p=vector a+2 times vector b-vector c, based on vectors a, b
Let vector c=m vector a+n vector b, then 2m-n=6,-4m+3n=5
M=23/2, n=17, i.e. vector c=(23/2) vector a+17 vector b.
Then vector p=vector a+2 vector b-vector c=vector a+2 vector b-[(23/2) vector a+17 vector b ]=(-21/2) vector a-15 vector b.
I hope you can accept my answer!
Known vector A =(λ,2), B=(-3,5), and vector A vs. If the included angle of b is an acute angle, the value range of λ is ______. Known vector A =(λ,2), B=(-3,5), and vector A and If the included angle of b is an acute angle, the value range of λ is ______.
From the meaning of the title
A•
B >0, and
A vs.
B is not collinear, i.e.-3 10>0, and λ
−3=2
5,
Solution (,−6
5)∪(−6
5,10
3),
Therefore, the answer is:(,−6
5)∪(−6
5,10
3).
From the meaning
A•
B >0, and
A and
B is not collinear, i.e.-3 10>0, and λ
−3=2
5,
Solution (,−6
5)∪(−6
5,10
3),
Therefore, the answer is:(,−6
5)∪(−6
5,10
3).
Given that a vector =(2,1), b vector =(m,6), and the acute angle between vector a and vector b, the value range of real number m is
6×1/2=3 6×2=12 So m >-3(m=12)
Given vector a=(3,4), vector b=(2, m), and the angle between a and b is acute, the value range of m is
The upstairs answer is wrong.
The angle between a and b is acute
A•b=3×2+4m=6+4m >0
Solution m >-3/2
And a and b are not collinear
3M-8=0
Solution m =8/3
The value range of m is (-3/2,8/3)∪(8/3,+∞)
The answer upstairs is wrong.
The angle between a and b is an acute angle.
A•b=3×2+4m=6+4m >0
Solution m >-3/2
And a and b are not collinear
3M-8=0
Solution m=8/3
The value range of m is (-3/2,8/3)∪(8/3,+∞)
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