The question is to find the distribution of $(\mathbf{a}^TZ, \mathbf{b}^TZ)$ where $Z$ is the standard $n$-variate normal random vector and $\mathbf{a}, \mathbf{b} \in \mathbb{R}^n$ are orthogonal.
I've calculated $\phi_Z(\mathbf{t}) = \exp(-\frac{1}{2}\mathbf{t}\mathbf{t}^T)$.
By definition, $\phi_Y(\mathbf{t}) = \mathbb{E}\exp[i(t_1 \mathbf{a}^{T}Z + t_2 \mathbf{b}^TZ)]$.
What should I do next?
- from this question, $\mathbf{a}^TZ$ and $\mathbf{b}^TZ$ are normally distributed, but as they're both functions of $Z$, I don't think they're independent
Let $$X\sim N(\boldsymbol\mu, \boldsymbol\Sigma)$$
Then, the affine transformation of $X$:$$Y=\mathbf{c}+\mathbf{B}X$$ follows the normal distribution:
$$N\left(\mathbf{c} + \mathbf{B} \boldsymbol\mu, \mathbf{B} \boldsymbol\Sigma \mathbf{B}^{\rm T}\right).$$
In your case,
$$\boldsymbol\mu=\boldsymbol0, \boldsymbol\Sigma=\boldsymbol I, \boldsymbol c=\boldsymbol0, \boldsymbol B=\begin{pmatrix} v_1 \\ v_n \end{pmatrix}.$$
Hence, $$\mathbf{B} \boldsymbol\Sigma \mathbf{B}^{\rm T}= \boldsymbol I.$$
The covariance matrix becomes the identity matrix, showing that $a^TZ$ and $b^TZ$ are independent.
You could directly compute the covariance of the two transformations as
$$\text{cov}(a^TZ, b^TZ)= a^T\boldsymbol I b=a^Tb=0.$$
As the joint distribution of them follows the bivariate normal distribution, they become independent if and only if their covariance is zero.