If the space vector m.n.p satisfies M = n, n = P, then M = P If n is a zero vector, the direction of the zero vector is indefinite

yes.
If n is a zero vector, then M and P are both zero vectors

Is the operation of vector scalar product suitable for multiplicative associative law? Why?

Not suitable. For example, a vector * (b vector * C vector) must first make it clear that the scalar product is a number. The previous example (b vector * C vector) represents a vector collinear with a, similar to μ a, because the scalar product is a number. Similarly, a vector * B vector * C vector represents a vector collinear with C

Why does the scalar product of a vector not satisfy the associative law?

From the formula of the law of association, (a · b) is a number, so the result of (a · b) · C is a vector with the same direction as C, while the direction of the vector calculated by a · (B · C) is the same as a, and the direction is different, so it does not satisfy the law of association

Prove the associative law in the operation law of vector scalar product
How do you prove it,
Please be more detailed

You have to draw a picture
Let OA = (a, b), ab = (C, d)
Since the same base is selected, so: (coordinate)
Points a (a, b), B (A-C, B + D)
Now let's consider the original definition of quantity product
(defined on the x-axis): ax = | a | cos ·| B | cos
So there is also ay = | a | cos ·| B | cos on the y-axis
The vector integral in the coordinate system is the sum of two parts: a · B = | a | cos ·| B | cos + | a | cos ·| B | cos
=a·c+b·d
∴(a,b)·(c,d)=ac+bd
The original reason can be proved
In addition, this theorem can also be used to judge vertical and equal conditions
This method is purely out of their own, not to look up information

If the space vector m.n.p satisfies M = n, n = P, then M = P If n is a zero vector, the direction of the zero vector is indefinite