Suppose that we have a random vector $\mathbf{v} \in \mathbb{R}^m$, where each element is sampled from a same distribution of variance $\sigma^2$.
Now, we have a constant vector $\mathbf{c} \in \mathbb{R}^m$ whose euclidian norm is $N$.
In this case why is the variance of inner product $\langle \mathbf{v}, \mathbf{c} \rangle$ equal to $\sigma^2 N^2$?
Let me use the matrix notation to make it easier to see. I also assume that the elements of $\mathbf v$ are independents hence the covariance matrix of $\mathbf v$ is $\sigma^2 I$. Observe that \begin{align*} \mathrm{Var}(\mathbf v^T \mathbf c) &= \mathbb E[(\mathbf v^T \mathbf c)^2]-\mathbb E[\mathbf v^T \mathbf c]^2\\ &= \mathbb E[(\mathbf v^T \mathbf c)^T(\mathbf v^T \mathbf c)]-\mathbb E[\mathbf v^T \mathbf c]^T\mathbb E[\mathbf v^T \mathbf c]\\ &= \mathbb E[\mathbf c^T\mathbf v\mathbf v^T \mathbf c]-\mathbb E[\mathbf c^T \mathbf v]\mathbb E[\mathbf v^T \mathbf c]\\ &= \mathbf c^T\mathbb E[\mathbf v\mathbf v^T ]\mathbf c-\mathbf c^T\mathbb E[ \mathbf v]\mathbb E[\mathbf v^T ]\mathbf c\\ &= \mathbf c^T\left(\mathbb E[\mathbf v\mathbf v^T ]-\mathbb E[ \mathbf v]\mathbb E[\mathbf v^T ]\right)\mathbf c\\ &= \mathbf c^T \mathrm{Var}(\mathbf v)\mathbf c\\ &= \mathbf c^T (\sigma^2 I) \mathbf c\\ &= \sigma^2 \mathbf c^T\mathbf c\\ &= \sigma^2 N^2\\ \end{align*}
On the second line, I can take the transpose since those are only real values, the transpose of a real value is itself. On fourth line I use linearity of expectation.
