Given the vector a=(1,1), b=(4, x), u=a+2b, v=2a+b, and u is parallel to v, then x=

Given the vector a=(1,1), b=(4, x), u=a+2b, v=2a+b, and u is parallel to v, then x=

U=a+2b
=(1+8,1+2X)
=(9,1+2X)
V =2a+b
=(2+4,2+X)
=(6,2+X)
U||v
9*(2+X)=(1+2x)*6
X =4

Vector a=(1,2), b=(x,1), when a+2b is perpendicular to 2a-b, find x?

A+2b=(1+2x,4)2a-b=(2-x,3). Since a+2b, is perpendicular to 2a-b, then (1+2x,4)×(2-x,3)=0(1+2x)×(2-x)+4×3=o x=7/2 or x=2

A+2b=(1+2x,4)2a-b=(2-x,3). Since a+2b, is perpendicular to 2a-b, then (1+2x,4)×(2-x,3)=0(1+2x)×(2-x)+4×3=o gives the solution x=7/2 or x=2

A+2b=(1+2x,4)2a-b=(2-x,3). Since a+2b, is perpendicular to 2a-b, then (1+2x,4)×(2-x,3)=0(1+2x)×(2-x)+4×3=o gives x=7/2 or x=2

A high school school mathematical vector problem,|a|=6,|b|=4,π/6,|2a-b||a-2b|

(2A-b)2=4|a|2-4|a||b|cosπ/6+|b|2=144-48+16=112,|2a-b|=4√7(a-2b)2=|a|2-4|a||b|cosπ/6+4|b|2=36-48+64=52,|a-2b|=2√13 2a -- b||a-2b|=4√7+2√13.

Given that ab is a non-zero vector,(2a-b)⊥(a+b)⊥(a-2b) then a, b has an angle θ equal to Given that ab is a non-zero vector,(2a-b)⊥(a+b)⊥(a-2b) then a, b with angle θ equals

(Both a and b below represent vectors, and θ represents the calculated angle)
As known,(a-2b)·a=0,
I. e.|a|^2-2|a||b|cosθ=0,
Similarly,|^2-2|a||b|cosθ=0,
Solving the two equations |a|=|b|,
Into one of the equations cos θ=1/2,
Therefore, the angle of the vector is 60°.

(Both a and b below represent vectors, and θ represents the calculated angle)
As known,(a-2b)·a=0,
I. e.|^2-2|a||b|cosθ=0,
Similarly,|^2-2|a||b|cosθ=0,
Solving the two equations |a|=|b|,
Into one of the equations cos θ=1/2,
Therefore, the angle of the vector is 60°.

(Both a and b below represent vectors, and θ represents the calculated angle)
As known,(a-2b)·a=0,
I. e.|a|^2-2|a||b|cosθ=0,
In the same way,|^2-2|a||b|cosθ=0,
Solving the two equations |a|=|b|,
Into one of the equations cos θ=1/2,
Therefore, the angle of the vector is 60°.

A vector =(1,2), b vector =(x,2), and a vector -2b vector is parallel to 2a vector - b vector, then x =?

A-2b=(1,2)-2(x,2)=(1-2x,-2)
2A-b=2(1,2)-(x,2)=(2-x,2)
A-2b‖2a-b
Inner product equals outer product
I.e.2(1-2x)=-2(2-x)
Solution x=1

Given vector a vector b,(2a+b)•(a-2b)=-(3√3)/2, find the angle between a and b

(2A+b)·(a-2b)=2||a|^2-2||b|^2-3a·b=2||a|^2-2||b|^2-3||a b|*cos
Shouldn't you tell the modulus of a and b? How do we do it?
If:|a|=|b|=1, then: cos = sqrt (3)/2, i.e.:=π/6

(2A+b)·(a-2b)=2||a|^2-2||b|^2-3a·b=2||a|^2-2||b|^2-3||a b|*cos
Shouldn't you tell the modulus of a and b? How do I do it?
If:|a|=|b|=1, then: cos = sqrt (3)/2, i.e.:=π/6