A judging problem of vector Let a, B and C be arbitrary non-zero plane vectors which are not collinear with each other 1.（a*b)*c-(c*a)*b=0 2.|a|-|b|＜|a-b| 3. (b * c) * a - (c * a) * B is not perpendicular to C 4.（3a+2b)*(3a-2b)=9|a|²-4|b|² Hope there is an explanation thank you

A judging problem of vector Let a, B and C be arbitrary non-zero plane vectors which are not collinear with each other 1.（a*b)*c-(c*a)*b=0 2.|a|-|b|＜|a-b| 3. (b * c) * a - (c * a) * B is not perpendicular to C 4.（3a+2b)*(3a-2b)=9|a|²-4|b|² Hope there is an explanation thank you

A. B and C. A are values with no direction, so it is impossible that the difference between two non collinear vectors is a zero vector. It can be seen that the scalar product of vectors does not satisfy the law of combination of multiplication. 2. Correct. Considering the relationship between the three sides of a triangle, the difference between the two sides is less than the third side