The vector a = (COSA, Sina), B = (- 1,2,3,2) proves that the vector a + B is perpendicular to a-b. when 2A + B = a-2b, the angle α is obtained

The vector a = (COSA, Sina), B = (- 1,2,3,2) proves that the vector a + B is perpendicular to a-b. when 2A + B = a-2b, the angle α is obtained

Vector a + B = (cosa-1 / 2, Sina + root 3 / 2)
Vector A-B = (COSA + 1 / 2, Sina radical 3-2)
(vector a + b) * (vector a-b), that is, the scalar product of vector a + B and vector A-B = the square of cosa - 1 / 4 + the square of Sina - 3 / 4 = the square of cosa + the square of Sina - 1 = 1-1 = 0
Because 2A + B = a-2b
So the square of 2A + B = the square of a-2b
That is, the square of 4 | a | plus the square of | B | plus 4 * a * B (vector) (quantitative product of vector a and vector b) = the square of a + the square of 4B - 4 * a * B (quantitative product) ①
|A | = √ (square of cosa + square of sina) = 1 | B | = √ (1 / 4 + 3 / 4) = 1 |
I don't know if α is the angle between a and B, if not, it's 90 degrees. If not, please make sure which angle α is. If you have any questions, please point out. I'm glad to answer for you