Problem 49 - Advanced Engineering Mathematics

Question

Find the area of the triangle determined by the given points. $$ P_{1}(1,2,4), P_{2}(1,-1,3), P_{3}(-1,-1,2) $$

Step-by-Step Solution

Answer

The area of the triangle is \(\sqrt{10}\).

Step 1: Understand the Formula for Area of a Triangle in 3D

To find the area of a triangle in 3D with vertices \(P_1(x_1, y_1, z_1)\), \(P_2(x_2, y_2, z_2)\), and \(P_3(x_3, y_3, z_3)\), we will use the formula for the area: \[\text{Area} = \frac{1}{2} \left| \mathbf{AB} \times \mathbf{AC} \right| \] where \(\mathbf{AB}\) and \(\mathbf{AC}\) are the vectors formed by the point coordinates.

Step 2: Calculate Vectors AB and AC

First, we calculate the vectors \(\mathbf{AB}\) and \(\mathbf{AC}\). These are determined by subtracting the coordinates of the points:- \(\mathbf{AB} = P_2 - P_1 = (1 - 1, -1 - 2, 3 - 4) = (0, -3, -1)\)- \(\mathbf{AC} = P_3 - P_1 = (-1 - 1, -1 - 2, 2 - 4) = (-2, -3, -2)\)

Step 3: Find the Cross Product of Vectors AB and AC

To find the cross product \(\mathbf{AB} \times \mathbf{AC}\), we set up the determinant:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \0 & -3 & -1 \-2 & -3 & -2 \\end{vmatrix}\]This results in:\[\mathbf{AB} \times \mathbf{AC} = (3 - 3)\mathbf{i} - ((0 - 2)\mathbf{j}) + (0 + 6)\mathbf{k} = (0, 2, 6)\]

Step 4: Calculate the Magnitude of the Cross Product

The magnitude of the cross product \(|\mathbf{AB} \times \mathbf{AC}|\) is:\[\sqrt{0^2 + 2^2 + 6^2} = \sqrt{0 + 4 + 36} = \sqrt{40}\]

Step 5: Calculate the Area of the Triangle

Finally, to find the area of the triangle using the formula from Step 1:\[\text{Area} = \frac{1}{2} \times \sqrt{40} = \frac{1}{2} \times 2\sqrt{10} = \sqrt{10}\]Thus, the area of the triangle is \(\sqrt{10}\).

Key Concepts

Cross Product

The cross product is an essential operation in vector algebra, especially useful in three-dimensional space when dealing with planes and surfaces. It is used to find a vector that is perpendicular to two given vectors. Let's say you have two vectors, \( \mathbf{A} \) and \( \mathbf{B} \), the cross product \( \mathbf{A} \times \mathbf{B} \) results in a vector that is orthogonal to both \( \mathbf{A} \) and \( \mathbf{B} \).
For our given problem, the vectors \( \mathbf{AB} = (0, -3, -1) \) and \( \mathbf{AC} = (-2, -3, -2) \) were crossed to find \( \mathbf{AB} \times \mathbf{AC} \). This was computed using a method similar to finding a determinant, resulting in the vector \( (0, 2, 6) \).
This new vector gives geometric insights. It tells us the direction perpendicular to the plane made by the two original vectors, and is crucial in determining vector area.

Vector Magnitude

Vector magnitude is a way of measuring the "length" or "size" of a vector in space. In three dimensions, the magnitude \( |\mathbf{V}| \) of a vector \( \mathbf{V} = (x, y, z) \) is calculated using the formula: \[|\mathbf{V}| = \sqrt{x^2 + y^2 + z^2}\]
This formula is the extension of the Pythagorean theorem into three-dimensional space. Magnitude gives us a scalar quantity that represents the length of the vector.
In our problem, after computing the cross product \( \mathbf{AB} \times \mathbf{AC} = (0, 2, 6) \), we calculated its magnitude to be \( \sqrt{40} \). This helps in further finding the area of the triangle because only this magnitude, not the direction, is required for area calculations.

Determinants

Determinants play a notable role in calculating cross products. A determinant is a scalar value that can be computed from the elements of a square matrix, and is denoted in a matrix using vertical bars, much like absolute value symbols.
For cross products, determinants help find a vector that is perpendicular to two other vectors. This is done by arranging the components of the vectors into a 3x3 matrix with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the top row. The calculation looks like this: \[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \0 & -3 & -1 \-2 & -3 & -2 \\end{vmatrix}\]
The determinant is then expanded to provide a new vector with components that combine the multiplication and subtraction of the determinants’ components. This is how we obtained the cross product \( (0, 2, 6) \). Determinants thus bridge algebra and geometry, aiding in area and vector calculations.

Geometry in Three Dimensions

Geometry in three-dimensional space adds depth to the usual plane geometry, giving a new perspective on shapes like triangles and circles. It allows for the representation of shapes in space where every point has an x, y, and z coordinate.
When computing areas, one must consider the spatial orientation and positioning of the shapes. Vectors and cross products provide mathematical tools to deal with such orientation and dimensions. For triangles, the orientation can be understood using vectors and their cross products.
The area of a triangle in 3D, as given by the formula \( \text{Area} = \frac{1}{2} \left| \mathbf{AB} \times \mathbf{AC} \right| \), signifies how calculation extends into three dimensions by incorporating vector arithmetic. By understanding these shapes and their properties in 3D, students can gain deeper insights into spatial relationships and measurements.