Problem 27 - Precalculus Mathematics for Calculus

Question

Find the area of \(\triangle P Q R\). $$P(6,0,0), Q(0,-6,0), R(0,0,-6)$$

Step-by-Step Solution

Answer

The area of \(\triangle PQR\) is 18 square units.

Step 1: Understand the given coordinates

The points provided are the vertices of the triangle \( \triangle PQR \) in three-dimensional space. The coordinates are \( P(6, 0, 0) \), \( Q(0, -6, 0) \), and \( R(0, 0, -6) \). These points lie on the coordinate axes, forming a triangle in 3D space.

Step 2: Use the formula for area of a triangle in 3D

We use the determinant-based formula for the area of a triangle given vertices in 3D space:\[\text{Area} = \frac{1}{2} \sqrt{(x_2y_3 + y_2z_3 + z_2x_3 - x_3y_2 - y_3z_2 - z_3x_2)^2}\]Here, \( x_1, y_1, z_1 \) correspond to point \( P \), and similarly for the other points.

Step 3: Substitute the coordinates into the formula

Substitute the coordinates into the formula where \( x_1 = 6, y_1 = 0, z_1 = 0 \); \( x_2 = 0, y_2 = -6, z_2 = 0 \); \( x_3 = 0, y_3 = 0, z_3 = -6 \).Calculate the determinant part:\[0 \cdot 0 + (-6) \cdot (-6) + 0 \cdot 0 - 0 \cdot (-6) - 0 \cdot 0 - (-6) \cdot 0 = 36\]The area formula then becomes:\[\text{Area} = \frac{1}{2} \sqrt{36^2} = \frac{1}{2} \times 36 = 18\]

Step 4: Conclusion

The area of the triangle \( \triangle PQR \) is \( 18 \) square units.

Key Concepts

Determinant-Based Formula

Calculating the area of a triangle in three-dimensional space is a bit different from typical 2D calculations. In 3D, we use the determinant-based formula. This formula helps to elegantly find the area of a triangle even when it's situated in 3D space, which can be tricky due to additional axis involvement. The formula is as follows:\[\text{Area} = \frac{1}{2} \sqrt{(x_2y_3 + y_2z_3 + z_2x_3 - x_3y_2 - y_3z_2 - z_3x_2)^2}\]The determinants help manage the relationships between points in 3D space. This formula accounts for each vertex of the triangle by effectively "flattening" the space for calculation purposes.By setting the formula with coordinates of each vertex, it effectively calculates the "projected" area on the 2D plane. Understanding this use of determinants helps demystify a complex geometric problem, providing a neat solution to 3D area calculations.

3-Dimensional Geometry

3-dimensional geometry takes the familiar concepts of shapes and extends them into a new dimension. In 2D, figures are defined by 2 coordinates (x and y), which makes computations like area and perimeter relatively straightforward. However, in 3D geometry, we introduce a third coordinate (z), adding depth to the shapes and making visualization a bit more challenging. In 3D space, triangles are not restricted to lying flat on a plane. They can be tilted or positioned anywhere within the three axes. This adds complexity when calculating measurements such as area because you're not just considering the base and height, but also how the shape orients itself in the space. Working with 3D geometry helps you visualize and solve more complex spatial problems, enhancing both your math and spatial reasoning skills. The real-world applications are vast, including fields like engineering, architecture, and computer graphics.

Triangle Vertices Coordinates

Every triangle, whether in 2D or 3D, is defined by its vertices – the corner points joining the sides of the triangle. When dealing with 3D triangles, each vertex has three coordinates, often noted as - \((x_1, y_1, z_1)\) for vertex 1- \((x_2, y_2, z_2)\) for vertex 2- \((x_3, y_3, z_3)\) for vertex 3These coordinates indicate the vertex positions along the x, y, and z axes. In the given problem, we have:- Point \(P\) at \((6, 0, 0)\)- Point \(Q\) at \((0, -6, 0)\)- Point \(R\) at \((0, 0, -6)\)These points lie directly on the x, y, and z axes respectively, creating a visible triangular shape in 3D space.Understanding how to interpret these coordinates is crucial for solving spatial problems. With practice, you'll gain insight into how these vertices detail the shape's structure, guiding calculations like our area problem above.