1. **State the problem:** Find the unit tangent vector $\mathbf{T}$, the principal unit normal vector $\mathbf{N}$, and the unit binormal vector $\mathbf{B}$ for the curve given by $$\mathbf{r}(t) = \sin t\,\mathbf{i} + \cos t\,\mathbf{j} + t\sqrt{3}\,\mathbf{k}.$$ 2. **Recall formulas and rules:** - The unit tangent vector is $$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}.$$ - The principal unit normal vector is $$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}.$$ - The unit binormal vector is $$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t).$$ 3. **Compute the derivative $\mathbf{r}'(t)$:** $$\mathbf{r}'(t) = \frac{d}{dt}(\sin t)\mathbf{i} + \frac{d}{dt}(\cos t)\mathbf{j} + \frac{d}{dt}(t\sqrt{3})\mathbf{k} = \cos t\,\mathbf{i} - \sin t\,\mathbf{j} + \sqrt{3}\,\mathbf{k}.$$ 4. **Compute the magnitude of $\mathbf{r}'(t)$:** $$\|\mathbf{r}'(t)\| = \sqrt{(\cos t)^2 + (-\sin t)^2 + (\sqrt{3})^2} = \sqrt{\cos^2 t + \sin^2 t + 3} = \sqrt{1 + 3} = 2.$$ 5. **Find the unit tangent vector $\mathbf{T}(t)$:** $$\mathbf{T}(t) = \frac{1}{2}(\cos t\,\mathbf{i} - \sin t\,\mathbf{j} + \sqrt{3}\,\mathbf{k}).$$ 6. **Compute the derivative of $\mathbf{T}(t)$:** $$\mathbf{T}'(t) = \frac{1}{2}(-\sin t\,\mathbf{i} - \cos t\,\mathbf{j} + 0\,\mathbf{k}) = \frac{1}{2}(-\sin t\,\mathbf{i} - \cos t\,\mathbf{j}).$$ 7. **Compute the magnitude of $\mathbf{T}'(t)$:** $$\|\mathbf{T}'(t)\| = \frac{1}{2} \sqrt{(-\sin t)^2 + (-\cos t)^2} = \frac{1}{2} \sqrt{\sin^2 t + \cos^2 t} = \frac{1}{2} \cdot 1 = \frac{1}{2}.$$ 8. **Find the principal unit normal vector $\mathbf{N}(t)$:** $$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} = \frac{\frac{1}{2}(-\sin t\,\mathbf{i} - \cos t\,\mathbf{j})}{\frac{1}{2}} = -\sin t\,\mathbf{i} - \cos t\,\mathbf{j}.$$ 9. **Find the unit binormal vector $\mathbf{B}(t)$ by cross product:** $$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \frac{1}{2}(\cos t, -\sin t, \sqrt{3}) \times (-\sin t, -\cos t, 0).$$ Calculate the cross product components: - $x = (-\sin t)(0) - (\sqrt{3})(-\cos t) = 0 + \sqrt{3} \cos t = \sqrt{3} \cos t$ - $y = (\sqrt{3})(-\sin t) - (\cos t)(0) = -\sqrt{3} \sin t$ - $z = (\cos t)(-\cos t) - (-\sin t)(-\sin t) = -\cos^2 t - \sin^2 t = -1$ So, $$\mathbf{B}(t) = \left(\sqrt{3} \cos t, -\sqrt{3} \sin t, -1\right).$$ 10. **Normalize $\mathbf{B}(t)$:** Magnitude: $$\sqrt{(\sqrt{3} \cos t)^2 + (-\sqrt{3} \sin t)^2 + (-1)^2} = \sqrt{3 \cos^2 t + 3 \sin^2 t + 1} = \sqrt{3(\cos^2 t + \sin^2 t) + 1} = \sqrt{3 + 1} = 2.$$ Unit binormal vector: $$\mathbf{B}(t) = \frac{1}{2}(\sqrt{3} \cos t, -\sqrt{3} \sin t, -1).$$ **Final answers:** $$\mathbf{T}(t) = \frac{1}{2}(\cos t, -\sin t, \sqrt{3}),$$ $$\mathbf{N}(t) = (-\sin t, -\cos t, 0),$$ $$\mathbf{B}(t) = \frac{1}{2}(\sqrt{3} \cos t, -\sqrt{3} \sin t, -1).$$