Verify that the points are the vertices of a parallelogram, and find its area.
A(1, 1, 3), B(0, −7, 1), C(2, −4, −6), D(3, 4, −4)


To verify that the points \( A(1, 1, 3) \), \( B(0, -7, 1) \), \( C(2, -4, -6) \), and \( D(3, 4, -4) \) are the vertices of a parallelogram, we can check if the opposite sides of the quadrilateral formed by these points are parallel. Then, we will compute the area of the parallelogram using the cross product of two adjacent sides.

### Step 1: Check if opposite sides are parallel
The opposite sides of a parallelogram must be parallel, meaning their direction vectors should be equal or scalar multiples of each other.

#### Find the direction vectors for opposite sides:
- Vector \( \overrightarrow{AB} = B - A = (0 - 1, -7 - 1, 1 - 3) = (-1, -8, -2) \)
- Vector \( \overrightarrow{CD} = D - C = (3 - 2, 4 - (-4), -4 - (-6)) = (1, 8, 2) \)

Since \( \overrightarrow{AB} = -\overrightarrow{CD} \), the vectors are scalar multiples of each other, meaning \( \overrightarrow{AB} \parallel \overrightarrow{CD} \).

- Vector \( \overrightarrow{AD} = D - A = (3 - 1, 4 - 1, -4 - 3) = (2, 3, -7) \)
- Vector \( \overrightarrow{BC} = C - B = (2 - 0, -4 - (-7), -6 - 1) = (2, 3, -7) \)

Since \( \overrightarrow{AD} = \overrightarrow{BC} \), the vectors are equal, meaning \( \overrightarrow{AD} \parallel \overrightarrow{BC} \).

Thus, opposite sides are parallel, confirming that the points form a parallelogram.

### Step 2: Find the area of the parallelogram
The area of a parallelogram can be found by taking the magnitude of the cross product of two adjacent sides.

We can use vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) for this purpose:
- \( \overrightarrow{AB} = (-1, -8, -2) \)
- \( \overrightarrow{AD} = (2, 3, -7) \)

The cross product of \( \overrightarrow{AB} \times \overrightarrow{AD} \) is:

\[
\overrightarrow{AB} \times \overrightarrow{AD} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & -8 & -2 \\
2 & 3 & -7
\end{vmatrix}
\]

Now, let's compute the determinant:

\[
\overrightarrow{AB} \times \overrightarrow{AD} = \mathbf{i} ((-8)(-7) - (-2)(3)) - \mathbf{j} ((-1)(-7) - (-2)(2)) + \mathbf{k} ((-1)(3) - (-8)(2))
\]

\[
= \mathbf{i} (56 - (-6)) - \mathbf{j} (7 - (-4)) + \mathbf{k} (-3 - (-16))
\]
\[
= \mathbf{i} (62) - \mathbf{j} (11) + \mathbf{k} (13)
\]
\[
= (62, -11, 13)
\]

The magnitude of the cross product gives the area of the parallelogram:
\[
\text{Area} = \sqrt{62^2 + (-11)^2 + 13^2} = \sqrt{3844 + 121 + 169} = \sqrt{4134} \approx 64.28
\]

Thus, the area of the parallelogram is approximately \( 64.28 \) square units.

### Summary:
1. The points form a parallelogram because opposite sides are parallel.
2. The area of the parallelogram is approximately \( 64.28 \) square units.

Would you like more details on any part of the solution, or have any questions?

Here are 5 related questions:
1. How do you find the volume of a parallelepiped using vectors?
2. What is the significance of the cross product in vector mathematics?
3. How can you verify if four points form a rectangle in 3D space?
4. What is the formula for the cross product of two vectors in 3D?
5. How can the dot product be used to find the angle between two vectors?

**Tip:** When checking for parallelism, remember that if two vectors are scalar multiples of each other, they are parallel.

Verify that the points are the vertices of a parallelogram, and find its area. A(1, 1, 3), B(0, −7, 1), C(2, −4, −6), D(3, 4, −4)

Verify that the points A(1, 1, 3), B(0, −7, 1), C(2, −4, −6), D(3, 4, −4) form a parallelogram by checking parallel sides and find the area using the cross product. Detailed steps with vector calculations and formulas provided. Suitable for Grades 11-12.

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