The nonzero vector a, b satisfies |a|=|b|=|a+b|, then the angle of a, b is? If the nonzero vectors a, b satisfy |a|=|b|=|a+b|, then the of a, b is?

The nonzero vector a, b satisfies |a|=|b|=|a+b|, then the angle of a, b is? If the nonzero vectors a, b satisfy |a|=|b|=|a+b|, then the of a, b is?

A|=|b|=|a+b|Vectors a, b are nonzero
Vectors a, b, a+b form an equilateral triangle (as shown)
The inner angle of an equilateral triangle is 60o
180O-60o =120o
 

If the non-zero vectors a and b satisfy |a|=|b|=|a-b|, then the angle between a and a+b is 60 degrees. Why? RT

1. For |a|=|b|=|a-b|, vectors a, b, a-b form an equilateral triangle.
2. The second vector a+b is the angular bisector of vectors a and b.
3. The angle between a and a+b is 30°.

Given vector a, vector b satisfies |a|=3,|b|=2|a+b|=4, then |a-b|= A Root 3 B Root 5 C3 D10 Given vector a, vector b satisfies |a|=3,|b|=2|a+b|=4, then |a-b|= A root 3 B root 5 C3 D10

|^2=||A+b|^2-4a*b
|A+b|^2=a^2+2ab+b^2=16
9+2Ab+4=16
2Ab=3
|^2=||A+b|^2-4ab=16-2*3=10
|A-b|=Root 10

|^2=||A+b|^2-4a*b
|A+b|^2=a^2+2ab+b^2=16
9+2Ab+4=16
2Ab=3
|A-b|^2=|a+b|^2-4ab=16-2*3=10
|A-b|=Root 10

Given vector a, b satisfies |a|=3|a+b=5|a-b|=5 find |b|

(A+b)2=a2+b2+2ab=25;
(A-b)2=a2+b2-2ab=25;
2A2+2b2=50;
B2=25-9=16;
|B |=4;

Given vector a+b+c=0,|a|=3,|b|=5,|c|=7, find the angle between a and b

A+b+c=0
A+b=-c
(A+b)2=|a|2+2ab+|b|2=|c|2
Ab=(|c|2-|a|2-|b|2)/2=15/2
A|b|cosα=15/2
Cos α=1/2
α=60°
I don't know how to welcome questions.

Given the non-zero vector a, b satisfies |a|=2,|b|=3,(a-2b)·(2a-b)=-1, find the included angle between a and b Given non-zero vector a, b satisfies |a|=2,|b|=3,(a-2b) point multiplication (2a-b)=-1, find the included angle between a and b

Let |a-b|=a2+b2-2|a||b|cosθ, a|=|b|=|a-b|. Let |a| replace |b| and |a-b| with |a|, then θ=60°a+b with cosθ=0.5, and the angle between a and a+b is 30°.

Let |a-b|=a2+b2-2|a||b|cosθ, a|=|b|=|a-b|. Let |a| replace |b| and |a-b| with |a|.