1. The problem asks to find the magnitude of the sum of two vectors $\tilde{q}$ and $\tilde{v}$, given that $\epsilon = |\tilde{q}|$, $1 = |\tilde{v}|$, and $\frac{\epsilon_1}{\epsilon} = \tilde{q} \tilde{v}$ (which we interpret as a relation involving the vectors or their magnitudes). 2. We know the magnitude of the sum of two vectors is given by the law of cosines: $$\sum = |\tilde{q} + \tilde{v}| = \sqrt{|\tilde{q}|^2 + |\tilde{v}|^2 + 2|\tilde{q}||\tilde{v}|\cos\theta}$$ where $\theta$ is the angle between $\tilde{q}$ and $\tilde{v}$. 3. Substitute the known magnitudes: $$|\tilde{q}| = \epsilon, \quad |\tilde{v}| = 1$$ so $$\sum = \sqrt{\epsilon^2 + 1 + 2\epsilon \cos\theta}$$ 4. To find $\cos\theta$, we use the given relation $\frac{\epsilon_1}{\epsilon} = \tilde{q} \tilde{v}$. Assuming this means the dot product normalized by $\epsilon$ equals $\epsilon_1$, then: $$\tilde{q} \cdot \tilde{v} = |\tilde{q}||\tilde{v}|\cos\theta = \epsilon \cdot 1 \cdot \cos\theta = \epsilon \cos\theta$$ Given $\frac{\epsilon_1}{\epsilon} = \tilde{q} \tilde{v}$, we interpret $\tilde{q} \tilde{v} = \epsilon_1 / \epsilon$, so: $$\epsilon \cos\theta = \frac{\epsilon_1}{\epsilon} \implies \cos\theta = \frac{\epsilon_1}{\epsilon^2}$$ 5. Substitute $\cos\theta$ back into the magnitude formula: $$\sum = \sqrt{\epsilon^2 + 1 + 2\epsilon \cdot \frac{\epsilon_1}{\epsilon^2}} = \sqrt{\epsilon^2 + 1 + 2 \frac{\epsilon_1}{\epsilon}}$$ 6. This is the expression for the magnitude of the sum of the vectors $\tilde{q}$ and $\tilde{v}$ in terms of $\epsilon$ and $\epsilon_1$. **Final answer:** $$\boxed{\sum = \sqrt{\epsilon^2 + 1 + 2 \frac{\epsilon_1}{\epsilon}}}$$