I'm reading Leif Andersen's "Interest Rate Modeling, vol 2 Term Structure Models" and met a problem on Chapter 14 LM Dynamics and Measures, $\S$ 14.2.5 Stochastic Volatility, Lemma 14.2.6, on page 602.
Before the lemma the stochastic volatility LMM model was defined in spot measure $Q^B$ (numeraire is the discrete banking account $B(t)=P(t,T_{q(t)}) \prod \limits _{n=0} ^{q(t)-1} \left(1+\tau_nL_n(T_n)\right)$).
$$dz(t) = \theta (z_0-z(t)) dt + \eta \psi(z(t)) dZ(t), \quad z(0) = z_0 \tag{14.15}$$ $$dL_n(t) = \sqrt{z(t)} \varphi(L_n(t)) \lambda_n(t)^\top \left(\sqrt{z(t)} \mu_n(t) dt + dW^B(t) \right) \tag{14.16}$$ where $$\mu_n(t) = \sum_{j=q(t)}^n \frac{\tau_j \varphi(L_j(t)) \lambda_j(t)}{1+\tau_j L_j(t)}$$ , and Z(t) a Brownian motion under the spot measure $Q^B$.
Now Lemma 14.2.6 says, the SDE for $z(t)$ in measure $Q^{T_{n+1}}$, $n\geq q(t) -1$ is
$$\begin{align} dz(t) &= \theta(z_0-z(t)) dt + \eta \psi(z(t)) \\ &\times \left( -\sqrt{z(t)} \mu_n(t)^\top \langle dZ(t), dW^B(t) \rangle + dZ^{n+1}(t) \right) \tag{14.17} \end{align} $$
, where $Z^{n+1}(t)$ is a Brownian motion in measure $Q^{T_{n+1}}$.
The proof on the book is as follows:
From earlier results, we have
$$dW^{n+1}(t) = \sqrt{z(t)} \mu_n(t) dt + dW^B(t)$$
Let us introduce the $m$-dimensional vector $$a(t) = \langle dZ(t), dW^B(t) \rangle / dt \tag{1}$$
so that we can write
$$dZ(t) = a(t)^\top dW^B(t) + \sqrt{1-\|a(t)\|^2}d\widetilde W(t) \tag{2}$$
where $\widetilde W(t)$ is a scalar Brownian motion independent of $W^B(t)$. In the measure $Q^{T_{n+1}}$, we then have
$$\begin{align} dZ(t) &= a(t)^\top \left(dW^{n+1}(t) - \sqrt{z(t)}\mu_n(t) dt \right) + \sqrt{1-\|a(t)\|^2} d\widetilde W(t) \tag{3}\\ &=dZ^{n+1}(t) - a(t)^\top \sqrt{z(t)} \mu_n(t) dt \tag{4} \end{align} $$ and the result follows.
My questions are:
Q1. Why with $(1)$ one can write $(2)$? is this a property for any 2 Brownian motions?
Q2. After $(3)$ I can only get $$dZ(t) = a(t)^\top dW^{n+1}(t) + \sqrt{1-\|a(t)\|^2}d\widetilde W(t) - a \sqrt{z(t)}\mu_n(t) dt$$
To get $(4)$ it means one can define Brownian motion $Z^{n+1}(t)$ in measure $Q^{T_{n+1}}$ as: $$dZ^{n+1} (t) = a(t)^\top dW^{n+1}(t) + \sqrt{1-\|a(t)\|^2}d\widetilde W(t)$$
How could one do so? notice $\widetilde W(t)$ is a Brownian motion in measure $Q^B$, but $ W^{n+1}(t)$ is a Brownian motion in measure $Q^{T_{n+1}}$, isn't here a bit messy?