Multiplying vector operators $ABC$ in quantum mechanics

Question

Given vector operators as $$\mathbf{A} = (A_{1}, A_{2} ,A_{3}) $$

$$\mathbf{B} = (B_{1}, B_{2} ,B_{3}) $$

$$\mathbf{C} = (C_{1}, C_{2} ,C_{3}) $$

I know that for two vector operators $$\begin{equation} \mathbf{Q} \mathbf{P} = \sum_{\alpha = 1}^{3} Q_{\alpha} P_{\alpha} \end{equation}$$

What is $$\mathbf{A}\mathbf{B}\mathbf{C}$$ in component form?

Edit: I'm looking for the product which appears as the result of quantisation rule which takes the classical expression to the quantum one.

Suppose the classical quantity we have are $\mathscr A_1(\mathbf{r,p,t})=\mathbf{r(p.r)}$,

$\mathscr A_2(\mathbf{r,p,t})=\mathbf{(r.p)r}$

What will be the quantum operator $\mathbf A_1$,$\mathbf A_2$ corresponding to $\mathscr A_1$ and $\mathscr A_2$?

Answer 1

Your expression is not really defined.

Your formula your have stated is the definition of the dot product.

Either your expression is

$$(\vec{A} \cdot \vec{B}) \vec{C}$$

In which case it is the vector $$(A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3}) \begin{bmatrix} C_{1}\\C_{2}\\C_{3}\end{bmatrix} = \begin{bmatrix}(A_{1}B_{1}C_{1} + A_{2}B_{2}C_{1} + A_{3}B_{3}C_{1}\\ (A_{1}B_{1}C_{2} + A_{2}B_{2}C_{2} + A_{3}B_{3}C_{2}\\A_{1}B_{1}C_{3} + A_{2}B_{2}C_{3} + A_{3}B_{3}C_{3} \end{bmatrix} $$

Or

$$\vec{A} (\vec{B} \cdot \vec{C})= (B_{1}C_{1} + B_{2}C_{2} + B_{3}C_{3}) \begin{bmatrix} A_{1}\\A_{2}\\A_{3}\end{bmatrix}$$

$$\begin{bmatrix}(B_{1}C_{1}A_{1} + B_{2}C_{2}A_{1} + B_{3}B_{3}A_{1}\\ B_{1}C_{1}A_{2} + B_{2}C_{2}A_{2} + B_{3}C_{3}A_{2}\\B_{1}C_{1}A_{3} + B_{2}C_{2}A_{3} + B_{3}C_{3}A_{3} \end{bmatrix} $$

Or as other answers have stated it is cross product