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To solve this problem, we are looking for a vector equation for the tangent line to the curve of intersection of the surfaces:

\[
z = x^2 + y^2 + 2y \quad \text{and} \quad x + z = 2
\]

at the point \( (1, 0, 1) \).

### Step 1: Understand the surfaces
The two surfaces are:
1. \( z = x^2 + y^2 + 2y \)
2. \( x + z = 2 \)

From the second equation, we can solve for \( z \):
\[
z = 2 - x
\]

### Step 2: Find the gradient of each surface
To find the tangent vector at the point, we need to compute the gradient vectors of both surfaces, as the tangent to the curve of intersection will be perpendicular to both gradients.

#### Gradient of surface 1 (\( z = x^2 + y^2 + 2y \)):

\[
\text{Let } F_1(x, y, z) = z - (x^2 + y^2 + 2y)
\]
\[
\nabla F_1 = \left( \frac{\partial F_1}{\partial x}, \frac{\partial F_1}{\partial y}, \frac{\partial F_1}{\partial z} \right)
= (-2x, -2y - 2, 1)
\]

At the point \( (1, 0, 1) \):
\[
\nabla F_1(1, 0, 1) = (-2(1), -2(0) - 2, 1) = (-2, -2, 1)
\]

#### Gradient of surface 2 (\( x + z = 2 \)):

\[
\text{Let } F_2(x, y, z) = x + z - 2
\]
\[
\nabla F_2 = \left( \frac{\partial F_2}{\partial x}, \frac{\partial F_2}{\partial y}, \frac{\partial F_2}{\partial z} \right)
= (1, 0, 1)
\]

At the point \( (1, 0, 1) \):
\[
\nabla F_2(1, 0, 1) = (1, 0, 1)
\]

### Step 3: Cross product of gradients
The direction of the tangent line is given by the cross product of the gradient vectors:

\[
\nabla F_1 \times \nabla F_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -2 & 1 \\ 1 & 0 & 1 \end{vmatrix}
= \hat{i} \left( (-2)(1) - (0)(1) \right) - \hat{j} \left( (-2)(1) - (1)(1) \right) + \hat{k} \left( (-2)(0) - (-2)(1) \right)
\]

\[
= \hat{i}(-2) - \hat{j}(-2 - 1) + \hat{k}(2)
= -2\hat{i} + 3\hat{j} + 2\hat{k}
\]

So, the direction vector of the tangent line is \( \langle -2, 3, 2 \rangle \).

### Step 4: Write the vector equation of the tangent line
The vector equation of the tangent line is:

\[
\mathbf{r}(t) = \mathbf{r_0} + t\mathbf{v}
\]
where \( \mathbf{r_0} = \langle 1, 0, 1 \rangle \) is the point of tangency, and \( \mathbf{v} = \langle -2, 3, 2 \rangle \) is the direction vector.

Thus, the parametric equation is:
\[
\mathbf{r}(t) = \langle 1, 0, 1 \rangle + t \langle -2, 3, 2 \rangle = \langle 1 - 2t, 3t, 1 + 2t \rangle
\]

### Step 5: Match with the given options
The correct parametric equation is:
\[
\langle 1 + 2t, -3t, 1 - 2t \rangle
\]
which corresponds to the selected answer.

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Would you like more details on any step? Here are 5 related questions to expand your understanding:

1. What is the significance of the cross product in this context?
2. How do gradients relate to tangent vectors?
3. How would the solution change if the point of tangency were different?
4. Can you explain how each partial derivative in the gradient is calculated?
5. What is the geometrical interpretation of a parametric equation of a line?

**Tip**: When dealing with surfaces, remember that the gradient points in the direction of the steepest ascent and is perpendicular to the tangent plane.

Find a vector equation for the tangent line to the curve of intersection of the surfaces z = x^2 + y^2 + 2y and x + z = 2 at the point (1, 0, 1).

Learn how to find the vector equation for the tangent line to the curve of intersection of the surfaces z = x^2 + y^2 + 2y and x + z = 2 at the point (1, 0, 1). This problem covers gradient vectors, cross product, and multivariable calculus.

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