Unless otherwise indicated all defined vectors lie on the surface of the unit sphere (i.e their norm is 1).
Let's take the following expression:
$(V\cdot Z_1)(V\cdot Z_2) = V^TZ_1V^TZ_2$
Given that the dot product is commutative the following is true:
$V^TZ_1V^TZ_2 = V^TZ_1 Z_2^TV$
Let's assume there exists a vector $Z'$ such that $V^TZ_1 Z_2^TV = V^TZ'$
Then:
$VV^TZ_1 Z_2^TV = VV^TZ'$ but since $V$ is of norm 1 we get:
$Z_1 Z_2^TV = Z'$
Which is a vector colinear to Z_1 and norm $Z_2^TV$.
However $Z_1$ and $Z_2$ were arbitrary so you can redo the proof and multiply by $V^T$ from the right instead and you get that $Z'$ must be colinear to $Z_2$. It is of course impossible for $Z'$ to always be colinear to 2 arbtirary vectors.
So one might be tempted to say no such $Z'$ exists. However that's not possible.
Take $0 \leq (V\cdot Z_1)(V\cdot Z_2) = \theta \leq 1$ Since that scalar is in the range of $\cos$, trivially there exists $Z'$ such that $Z' \cdot V = \theta$
What did I do wrong?
Even if $$V^T V = \left[ \begin{matrix} V_x & V_y & V_z \end{matrix} \right] \left[ \begin{matrix} V_x \\ V_y \\ V_z \end{matrix} \right] = V_x^2 + V_y^2 + V_z^2 = 1$$ you have $$V V^T = \left[ \begin{matrix} V_x \\ V_y \\ V_z \end{matrix} \right] \left[ \begin{matrix} V_x & V_y & V_z \end{matrix} \right] = \left[ \begin{matrix} V_x^2 & V_x V_y & V_x V_z \\ V_y V_x & V_y^2 & V_y V_z \\ V_z V_x & V_z V_y & V_z^2 \end{matrix} \right] \ne I$$ In particular, $$V V^T Z = \left[ \begin{matrix} V_x^2 & V_x V_y & V_x V_z \\ V_y V_x & V_y^2 & V_y V_z \\ V_z V_x & V_z V_y & V_z^2 \end{matrix} \right] \left[ \begin{matrix} Z_x \\ Z_y \\ Z_z \end{matrix} \right ] = \left[ \begin{matrix} V_x^2 Z_x + V_x V_y Z_y + V_x V_z Z_z \\ V_x V_y Z_x + V_y^2 Z_y + V_y V_z Z_z \\ V_x V_z Z_x + V_y V_z Z_y + V_z^2 Z_z \end{matrix} \right] \ne \left[ \begin{matrix} Z_x \\ Z_y \\ Z_z \end{matrix} \right ]$$ Perhaps you forgot that matrix multiplication isn't commutative?