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Given that the vector a, b has an angle of 60° and |a|=2,|b|=1, if (2a+b)⊥(ma-b), then the value of m? I find | Do I need this step to find the length of |2a+b|? And I don't know how to use it.
You don't have to do this. Just open it.
(2A+b)⊥(ma-b),(2a+b)·(ma-b)=0, disassembled
2Ma^2+(m-2) ab-b^2=0
|^2M |^2+(m-2) ab-|^2=0,
|^2M|^2+(m-2)||a||b||cos60°-|^2=0
|A|=2,|b|=1, m=1/3
Given the coordinates of point AB as (-1,0),(1,4), a vector =(2k+3,2), if a//AB vector, then k is
AB =(1-(-1),4-0)=(2,4)
Because AB is parallel to a, so
(2K+3)/2=2/4
2K+3=1, k=-1
Given the point A (-1,0), B (1,3), vector A =(2K-1,2), if vector AB is vertical vector A, find the value of K Given the point A (-1,0), B (1,3), vector A =(2K-1,2), if the vector AB vertical vector A, find the value of K
Vector AB=(2,3)
Vector AB is perpendicular to vector A, then vector AB*vector A=0
I.e.(2,3)·(2k-1,2)=0
2(2K-1)+3*2=0
4K-2+6=0
I.e. k=-1
Given that the coordinates of A and B are (2,-2) and (4,3) respectively, the coordinates of vector P are (2k-1,7), and P‖ vector AB, then the value of K is? Such as title
AB vector =(2,5)
The two vectors are parallel, and the outer product is zero:5(2k-1)-2*7=0
5(2K-1)=2*7
Solution k=19/10
A point (2k-1,7) is parallel to a vector AB (2,5) to find K One point (2k-1,7) is parallel to one vector AB (2,5) to find K
Is the line between the coordinate origin O and this point parallel to AB?
If so, the ratio of the corresponding coordinates is equal (more equivalent is the cross product equal)
That is,(2k-1)/2=7/5, k=1.9.
Welcome to ask ~
Vector a is known, vector b is two non-parallel vectors, and the values of the real numbers m, n satisfy the following conditions respectively 3 Vector a+4 vector b=(m-1) vector a+(20-n) vector b Vector a is known, vector b is two non-parallel vectors, and the values of the real numbers m, n satisfying the following conditions are obtained respectively 3 Vector a+4 vector b =(m-1) vector a+(20-n) vector b Vector a is known, vector b is two non-parallel vectors, and the values of the real numbers m, n satisfying the following conditions are obtained respectively 3 Vector a+4 vector b=(m-1) vector a+(20-n) vector b
The left and right of the equation, a, b, the equal,
3= M-1
4=20-N
So,
M=4
N=16
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