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Mathematics elective course 2- 1P98 1 1 explanation
Let the coordinates of vector P under the bases A+B, A-B and C be (x, y, z), then vector P = x (a+b)+y (a-b)+ZC = (x+y) A+(x-y) b+ZC.

And the coordinates of vector P in cardinal numbers A, B and C are (1, 2,3), so vector P=a+2b+3c. So there are:

X+y= 1, x-y=2, z=3, the solution is: X = 3/2, Y =- 1/2, Z = 3. Therefore, the coordinates of vector P under bases a+b, a-b and C are (3/2,-).

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