If vector A and vector B satisfy that the module of vector A is equal to 1, the module of vector B is equal to 2, and the module of vector A minus B is equal to 2, then the module of vector A plus B is obtained.

Solution 1: Draw a parallelogram and use the cosine theorem.
Solution 2:
(A-b)^2=a^2+b^2-2abcosC=4
2AbcosC=1
(A+b)^2= a^2+b^2+2abcosC=6
|A+b|= Root 6

What is the square of the absolute value of a vector minus b vector

The nonzero vector a⊥b, a-b equals?

The value of a-b, which is not equal to 0 a⊥b=0(a, b and 0 on the left are all vectors and should be written by hand → above), shall be determined as the case may be.

The module of vector a is not equal to zero, and vector a is not zero. A Sufficient Unnecessary B Necessary Inadequate C necessary and sufficient condition D is neither sufficient nor necessary condition I think if A is 6+6i (i is the unit of imaginary number), it is not zero vector, but module is zero. The module of vector a is not equal to zero, and vector a is not zero. A Sufficient Unnecessary B Necessary Inadequate C Sufficient and Necessary Condition D Inadequate and Unnecessary Condition I think if A is 6+6i (i is the unit of imaginary number), it is not zero vector, but the module is zero. The module of vector a is not equal to zero. Vector a is not zero. A sufficient not necessary B necessary not sufficient C necessary and sufficient condition D is neither sufficient nor necessary condition I think if A is 6+6i (i is the unit of imaginary number), it is not zero vector, but the module is zero.

6+6I module is under root number (6*6+6*6)=6 times root number 2

If the non-zero vector a b satisfies a minus b modulus is equal to a plus b modulus, then a b satisfies Same in direction A B direction opposite C modulus equal D is perpendicular to each other Given that the non-zero vector a b satisfies a minus b is equal to a plus b module, then a b satisfies Same in direction A B direction opposite C modulus equal D is perpendicular to each other

Vector problem ~ The modules of known non-zero vectors a, b.a are equal to the modules of 2 times b, and b is perpendicular (a+b) Find the included angle of ab

Because b is vertical (a+b), b (a+b)=0, the vector ab+b2=0(this expression must also be a vector)
So |a||b|cos b||b|=0, put |a||=2|b| in to eliminate other letters.
Cosθ=-1/2, included angle 120°

Because b is vertical (a+b), b (a+b)=0, the vector ab+b2=0(this expression must also be a vector)
So |a||b|cos b||b|=0, put |a||=2|b| in, you can eliminate other letters, and get
Cosθ=-1/2, included angle 120°

Because b is vertical (a+b), b (a+b)=0, the vector ab+b2=0(this expression must also be a vector)
So |a||b|cos b||b|=0, take |a||=2|b| and substitute it in to eliminate other letters.
Cosθ=-1/2, included angle 120°

If vector A and vector B satisfy that the module of vector A is equal to 1, the module of vector B is equal to 2, and the module of vector A minus B is equal to 2, then the module of vector A plus B is obtained.