To find the area of the quadrilateral ABCD with vertices A(0,4,1), B(2,3,-1), C(4,5,0), and D(2,6,2), we can use the formula for the area of a quadrilateral formed by two triangles. We will divide the quadrilateral into two triangles, ABC and ACD, and then calculate the area of each triangle using the cross product of vectors.
### Step-by-Step Solution:
1. **Find the vectors AB and AC**:
- The vector AB is calculated as:
\[
\text{AB} = B - A = (2, 3, -1) - (0, 4, 1) = (2 - 0, 3 - 4, -1 - 1) = (2, -1, -2)
\]
- The vector AC is calculated as:
\[
\text{AC} = C - A = (4, 5, 0) - (0, 4, 1) = (4 - 0, 5 - 4, 0 - 1) = (4, 1, -1)
\]
2. **Calculate the cross product AB × AC**:
- Using the determinant method for the cross product:
\[
\text{AB} \times \text{AC} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -1 & -2 \\
4 & 1 & -1
\end{vmatrix}
\]
- Expanding the determinant:
\[
= \mathbf{i} \left((-1)(-1) - (-2)(1)\right) - \mathbf{j} \left(2(-1) - (-2)(4)\right) + \mathbf{k} \left(2(1) - (-1)(4)\right)
\]
\[
= \mathbf{i} (1 + 2) - \mathbf{j} (-2 + 8) + \mathbf{k} (2 + 4)
\]
\[
= 3\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}
\]
3. **Magnitude of the cross product**:
- The magnitude of the vector \( \mathbf{v} = (3, -6, 6) \) is calculated as:
\[
|\mathbf{v}| = \sqrt{3^2 + (-6)^2 + 6^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9
\]
4. **Area of triangle ABC**:
- The area of triangle ABC is given by:
\[
\text{Area}_{ABC} = \frac{1}{2} |\mathbf{AB} \times \mathbf{AC}| = \frac{1}{2} \times 9 = 4.5
\]
5. **Find the vectors AD and AC**:
- The vector AD is calculated as:
\[
\text{AD} = D - A = (2, 6, 2) - (0, 4, 1) = (2, 2, 1)
\]
6. **Calculate the cross product AD × AC**:
- Using the determinant method for the cross product:
\[
\text{AD} \times \text{AC} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 2 & 1 \\
4 & 1 & -1
\end{vmatrix}
\]
- Expanding the determinant:
\[
= \mathbf{i} \left(2(-1) - 1(1)\right) - \mathbf{j} \left(2(-1) - 1(4)\right) + \mathbf{k} \left(2(1) - 2(4)\right)
\]
\[
= \mathbf{i} (-2 - 1) - \mathbf{j} (-2 - 4) + \mathbf{k} (2 - 8)
\]
\[
= -3\mathbf{i} + 6\mathbf{j} - 6\mathbf{k}
\]
7. **Magnitude of the cross product**:
- The magnitude of the vector \( \mathbf{w} = (-3, 6, -6) \) is calculated as:
\[
|\mathbf{w}| = \sqrt{(-3)^2 + 6^2 + (-6)^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9
\]
8. **Area of triangle ACD**:
- The area of triangle ACD is given by:
\[
\text{Area}_{ACD} = \frac{1}{2} |\mathbf{AD} \times \mathbf{AC}| = \frac{1}{2} \times 9 = 4.5
\]
9. **Total Area of Quadrilateral ABCD**:
- The total area of quadrilateral ABCD is:
\[
\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = 4.5 + 4.5 = 9
\]
### Final Answer:
The area of the quadrilateral ABCD is \( 9 \) square units.
