Given vectors a=(1,2), b=(2,-1), then the angle between vector a and vector b

Given vectors a=(1,2), b=(2,-1), then the angle between vector a and vector b

Let the angle be α
Then cosα=[1*2+2*(-1)]/√(12+22)[22+(-1)2]=0
So α=π/2

Vector Problem: Given |a|=1,|b|=2, a·b=1, then (a+b)·(a-b) angle

(A+b).(a-b)=a^2-b^2.=1-2^2.=1-4.=-3.|a+b|=√(a+b)^2.=√(a^2+2ab+b^2).=√(1^2+2*1+2^2). a+b|=√7. Similarly,|a-b|=√3.cos <(a+b),(a-b)>=(a+b).(a-b)/|a+b||a-b|.=-3/(√7 3).=-3√21/21.<(a+b),(a...

If vector a and vector b satisfy vector a|= vector b|=1, and vector a• vector b+ vector b• vector b=3/2, The angle between vector a and vector b is ()

|A|=|b|=1
A•b+b•b=3/2
So a•b=3/2-|b|2=1/2
So |a b|*cosθ=1/2
So 1*1*cosθ=1/2
I.e. cos θ=1/2
Therefore, the included angle between vector a and vector b is (60°)
If don't understand, wish study happy!

Given vector a, b satisfies |a+b|=8,|a-b|=6, find a•b Given vectors a, b satisfy |a+b|=8,|a-b|=6, find a•b

|A+b|=8=>(a+b)2=64=>
A2+ b2+2ab =64 1
|A-b|=6=>(a-b)2=36=>
A2+ b2-2ab =36 2
①-②4Ab=28 ab=7

|A+b|=8=>(a+b)2=64=>
A2+B2+2ab=64 1
|A-b|=6=>(a-b)2=36=>
A2+B2-2ab=36 2
①-②4Ab=28ab=7

Given the vector |a|=2,|b|=3,(a, b)=60°, then a•b=what And why?

According to the formula of vector multiplication, it is equal to the product of the modulus of two vectors and the cosine value of the included angle
Ab=|a||b|cos60°=2x3x1/2=3

Given vector a=(2,3) b=(4,-2), then a•b=

This is a vector dot multiplication, defined a.b=2*4+3*(-2)=2