If vector a and B satisfy | vector a | = | vector B | = 1, and vector A &; vector B + vector B &; vector b = 3 / 2, Then the angle between vector a and vector B is ()
|a|=|b|=1
a•b+b•b=3/2
So a &; b = 3 / 2 - | B | 178; = 1 / 2
So | a | * | B | * cos θ = 1 / 2
So 1 * 1 * cos θ = 1 / 2
That is cos θ = 1 / 2
So the angle between vector a and vector B is (60 °)
If you don't understand, I wish you a happy study!
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