Sketch the triangle with vertices at the origin O, 
P = (7, 7, 0),
 and 
Q = (0, 7, 7),
 and compute its area using cross products.

To sketch the triangle and compute its area using cross products, we can proceed step-by-step.

### 1. **Vertices of the Triangle**
The triangle has vertices at:
- \( O = (0, 0, 0) \)
- \( P = (7, 7, 0) \)
- \( Q = (0, 7, 7) \)

### 2. **Vectors for the Triangle**
We need to find vectors \( \overrightarrow{OP} \) and \( \overrightarrow{OQ} \) based on the vertices:

- \( \overrightarrow{OP} = P - O = (7, 7, 0) \)
- \( \overrightarrow{OQ} = Q - O = (0, 7, 7) \)

### 3. **Cross Product of Vectors \( \overrightarrow{OP} \) and \( \overrightarrow{OQ} \)**
The area of the triangle can be calculated using the formula for the magnitude of the cross product of two vectors:

\[
\text{Area} = \frac{1}{2} \left\| \overrightarrow{OP} \times \overrightarrow{OQ} \right\|
\]

Let's calculate \( \overrightarrow{OP} \times \overrightarrow{OQ} \).

\[
\overrightarrow{OP} \times \overrightarrow{OQ} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
7 & 7 & 0 \\
0 & 7 & 7
\end{vmatrix}
\]

Expanding this determinant:

\[
\overrightarrow{OP} \times \overrightarrow{OQ} = \mathbf{i}(7 \cdot 7 - 0 \cdot 7) - \mathbf{j}(7 \cdot 7 - 0 \cdot 0) + \mathbf{k}(7 \cdot 7 - 7 \cdot 0)
\]

\[
\overrightarrow{OP} \times \overrightarrow{OQ} = \mathbf{i}(49) - \mathbf{j}(49) + \mathbf{k}(49)
\]

\[
\overrightarrow{OP} \times \overrightarrow{OQ} = (49, -49, 49)
\]

### 4. **Magnitude of the Cross Product**
The magnitude of \( \overrightarrow{OP} \times \overrightarrow{OQ} \) is calculated as:

\[
\left\| \overrightarrow{OP} \times \overrightarrow{OQ} \right\| = \sqrt{49^2 + (-49)^2 + 49^2}
\]

\[
\left\| \overrightarrow{OP} \times \overrightarrow{OQ} \right\| = \sqrt{3 \times 49^2}
\]

\[
\left\| \overrightarrow{OP} \times \overrightarrow{OQ} \right\| = \sqrt{3 \times 2401} = \sqrt{7203} = 84.85 \, (\text{approx})
\]

### 5. **Area of the Triangle**
The area of the triangle is half the magnitude of the cross product:

\[
\text{Area} = \frac{1}{2} \times 84.85 \approx 42.425
\]

So, the area of the triangle is approximately \( 42.43 \) square units.

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### Related Questions
1. How can the cross product of two vectors be used to find the area of a parallelogram?
2. What is the significance of the magnitude of a cross product in 3D geometry?
3. How does the determinant relate to the cross product in vector calculations?
4. What are some real-world applications of finding the area using cross products?
5. How can you verify the calculation of a cross product?

**Tip:** In 3D geometry, the cross product of two vectors not only gives the magnitude but also a vector perpendicular to the plane containing the original vectors.

Sketch the triangle with vertices at the origin O, P = (7, 7, 0), and Q = (0, 7, 7), and compute its area using cross products.

Learn how to compute the area of a triangle with vertices O(0,0,0), P(7,7,0), and Q(0,7,7) in 3D space using vector cross products. This problem involves vector calculations and cross products, suitable for students in Grades 10-12.

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