1. **State the problem:** Find the length of the curve given by the vector function $$\mathbf{r}(t) = (\sqrt{2}t, e^t, e^{-t})$$ for $$0 \leq t \leq 1$$. 2. **Formula for curve length:** The length $$L$$ of a curve defined by $$\mathbf{r}(t)$$ from $$t=a$$ to $$t=b$$ is given by: $$ L = \int_a^b \|\mathbf{r}'(t)\| \, dt $$ where $$\mathbf{r}'(t)$$ is the derivative of $$\mathbf{r}(t)$$ and $$\|\mathbf{r}'(t)\|$$ is its magnitude. 3. **Find the derivative:** $$ \mathbf{r}'(t) = \left( \frac{d}{dt}(\sqrt{2}t), \frac{d}{dt}(e^t), \frac{d}{dt}(e^{-t}) \right) = (\sqrt{2}, e^t, -e^{-t}) $$ 4. **Find the magnitude:** $$ \|\mathbf{r}'(t)\| = \sqrt{(\sqrt{2})^2 + (e^t)^2 + (-e^{-t})^2} = \sqrt{2 + e^{2t} + e^{-2t}} $$ 5. **Simplify the expression inside the square root:** Recall that $$e^{2t} + e^{-2t} = 2\cosh(2t)$$, so: $$ \|\mathbf{r}'(t)\| = \sqrt{2 + 2\cosh(2t)} = \sqrt{2(1 + \cosh(2t))} $$ 6. **Use the identity:** $$1 + \cosh(2t) = 2\cosh^2(t)$$, so: $$ \|\mathbf{r}'(t)\| = \sqrt{2 \cdot 2 \cosh^2(t)} = \sqrt{4 \cosh^2(t)} = 2\cosh(t) $$ 7. **Set up the integral for length:** $$ L = \int_0^1 2\cosh(t) \, dt $$ 8. **Integrate:** $$ \int 2\cosh(t) \, dt = 2\sinh(t) + C $$ 9. **Evaluate definite integral:** $$ L = 2\sinh(1) - 2\sinh(0) = 2\sinh(1) - 0 = 2\sinh(1) $$ 10. **Final answer:** $$ \boxed{L = 2\sinh(1)} $$ This is the length of the curve from $$t=0$$ to $$t=1$$.