The problem asks to find the symmetric equations of the tangent line to the curve defined by the vector function $\vec{r}(t) = 2 \cos(t) \hat{i} + 6 \sin(t) \hat{j} + t \hat{k}$ at $t = \frac{\pi}{3}$.

1. Problem Description

The problem asks to find the symmetric equations of the tangent line to the curve defined by the vector function $\vec{r}(t) = 2 \cos(t) \hat{i} + 6 \sin(t) \hat{j} + t \hat{k}$ at $t = \frac{\pi}{3}$.

2. Solution Steps

First, we need to find the derivative of $\vec{r}(t)$ with respect to $t$, which represents the tangent vector to the curve.

$\vec{r}'(t) = \frac{d}{dt}(2 \cos(t) \hat{i} + 6 \sin(t) \hat{j} + t \hat{k})$

$\vec{r}'(t) = -2 \sin(t) \hat{i} + 6 \cos(t) \hat{j} + 1 \hat{k}$

Next, we evaluate the tangent vector at $t = \frac{\pi}{3}$:

$\vec{r}'(\frac{\pi}{3}) = -2 \sin(\frac{\pi}{3}) \hat{i} + 6 \cos(\frac{\pi}{3}) \hat{j} + 1 \hat{k}$

Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$,

$\vec{r}'(\frac{\pi}{3}) = -2(\frac{\sqrt{3}}{2}) \hat{i} + 6(\frac{1}{2}) \hat{j} + 1 \hat{k}$

$\vec{r}'(\frac{\pi}{3}) = -\sqrt{3} \hat{i} + 3 \hat{j} + 1 \hat{k}$

Now, we need to find the position vector $\vec{r}(\frac{\pi}{3})$:

$\vec{r}(\frac{\pi}{3}) = 2 \cos(\frac{\pi}{3}) \hat{i} + 6 \sin(\frac{\pi}{3}) \hat{j} + \frac{\pi}{3} \hat{k}$

$\vec{r}(\frac{\pi}{3}) = 2(\frac{1}{2}) \hat{i} + 6(\frac{\sqrt{3}}{2}) \hat{j} + \frac{\pi}{3} \hat{k}$

$\vec{r}(\frac{\pi}{3}) = 1 \hat{i} + 3\sqrt{3} \hat{j} + \frac{\pi}{3} \hat{k}$

Thus, the point on the curve is $(1, 3\sqrt{3}, \frac{\pi}{3})$.

The tangent line has the direction vector $\vec{r}'(\frac{\pi}{3}) = <-\sqrt{3}, 3, 1>$. The parametric equations of the tangent line are given by:

$x = 1 - \sqrt{3}t$

$y = 3\sqrt{3} + 3t$

$z = \frac{\pi}{3} + t$

To find the symmetric equations, we solve for $t$ in each equation:

$t = \frac{x - 1}{-\sqrt{3}}$

$t = \frac{y - 3\sqrt{3}}{3}$

$t = z - \frac{\pi}{3}$

Therefore, the symmetric equations are:

$\frac{x - 1}{-\sqrt{3}} = \frac{y - 3\sqrt{3}}{3} = z - \frac{\pi}{3}$

3. Final Answer

$\frac{x-1}{-\sqrt{3}} = \frac{y-3\sqrt{3}}{3} = z-\frac{\pi}{3}$