The vector a=(sinx,3/2), b=(cosx,-1) is known. (1) When a//b, find the value of the square x-2sinxcosx of 2*cosx; (2) Find the minimum value of f (x)=2sinx+(vector a+vector b)·(vector a-vector b) on [-π/2,0] and the value of x when the minimum value is obtained. The vector a=(sinx,3/2), b=(cosx,-1). (1) When a//b, find the value of the square x-2sinxcosx of 2*cosx; (2) Find the minimum value of f (x)=2sinx+(vector a+vector b)·(vector a-vector b) on [-π/2,0] and the value of x when the minimum value is obtained.

The vector a=(sinx,3/2), b=(cosx,-1) is known. (1) When a//b, find the value of the square x-2sinxcosx of 2*cosx; (2) Find the minimum value of f (x)=2sinx+(vector a+vector b)·(vector a-vector b) on [-π/2,0] and the value of x when the minimum value is obtained. The vector a=(sinx,3/2), b=(cosx,-1). (1) When a//b, find the value of the square x-2sinxcosx of 2*cosx; (2) Find the minimum value of f (x)=2sinx+(vector a+vector b)·(vector a-vector b) on [-π/2,0] and the value of x when the minimum value is obtained.

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Given vector a=(1,2), vector b=(3,4) 1,2 Vector a vector a * vector b 2. If vector c=(-1,0), and (m vector c+vector b) is parallel to vector a, find the value of m. Given vector a=(1,2), vector b=(3,4) 1,2 Vector a vector a * vector b 2, If vector c=(-1,0), and (m vector c+vector b) is parallel to vector a, find the value of m. Given vector a=(1,2), vector b=(3,4) 1,2 Vector a vector a* vector b 2. If vector c=(-1,0), and (m vector c+vector b) is parallel to vector a, find the value of m.

1.2 Vector a=2(1,2)=(2,4)
Vector a* Vector b=1*3+2*4=3+8=11
2.M vector c + vector b = m (-1,0)+(3,4)=(-m,0)+(3,4)=(3-m,4)
M vector c + vector b shall be parallel to vector a, with relation (3-m)*2=1*4
Solution m=1

Given two vectors a, b satisfies |a|=1|b|=2a perpendicular a+b to find the angle between vectors a and b

Because a is vertical a+b, i.e.|a|^2+ vector a* vector b=0, vector a* vector b=-1
Cos = vector a* vector b/(|a b|)=-1/2
Angle 120°

Because a is vertical a+b, i.e.|a|^2+ vector a* vector b=0, vector a* vector b=-1
Cos = vector a* vector b/(|a b|)=-1/2
Included angle is 120°

Given that the vector ab satisfies |a|=2,|b|=1, and a•(a+b)=3, then a, b=? Given that the vector a b satisfies |a|=2,|b|=1, and a•(a+b)=3, then a, b=?

Solution
A (a+b)=3
A2+ ab =3
I.e.4+ab=3
Ab=-1
Also ab=/a//b/cos
/A//b/cos=-1
Cos=-1/(2×1)=-1/2
∵∈[0,π]
∴=120

Given vector a, b satisfies a*b=0,|a|=1,|b|=2, then |2a+b|=? I first find a=1, b=2 according to |a|^2=a^2, and then bring in |2a+b|. Why not? Given vector a, b satisfies a*b=0,|a|=1,|b|=2, then |2a+b|=? I first find a=1, b=2 according to |a |^2=a^2, and then bring in |2a+b|. Why not?

Correct one of your concepts first and suggest you don't write: a^2Because vectors do n' t have a square operation, so a^2 is: a^2=a ·a=|a |^2, the previous one: a^2-b^2=|a |^2-|b |^2 is yours, right? That's what's going on:|2a+b |^2=(2a+b)·(2a+b)=4|a |^2|...

Correct one of your concepts first and suggest you don't write: a^2Because vectors do n' t have a square operation, the so-called a^2 is: a^2=a·a==|a |^2, the previous one: a^2-b^2=|a |^2-|b |^2 is yours, which is the same thing:|2a+b |^2=(2a+b)·(2a+b)=4|a |^2|...

First correct one of your concepts, and suggest not to write: a^2 because the vector is not square operation, so-called a^2, in fact, is: a^2=a ·a=|a |^2, the first: a^2-b^2=|a |^2-|^2 is yours, this is the same thing:|2a+b |^2=(2a+b)·(2a+b)=4|a |^2|...

Given vector a, b satisfies |a|=2,|b|=1,(b-2a)⊥b, then |a+b|=

(B-2a)⊥b
Then:
(B-2a)*b=0
|B|2-2a*b=0
Get:
2A*b=|b|=1
Also:
|A+b|2=|a|2+2a*b+|b|2=2 2+1+1=6
Get:
|A+b|=√6