Question

In Exercises 27-30, find the area of the triangle with the given vertices. Vertices: (0,0,0),(1,3,-1) and (2,1,1) .

Short answer

The area of the triangle is approximately 3.54 square units.

Step-by-step solution

Understand the coordinates of the triangle's vertices

The given vertices of the triangle are points in a three-dimensional space: \((0,0,0)\), \((1,3,-1)\), and \((2,1,1)\).

Use vectors to define two sides of the triangle

We can define two vectors that represent two sides of the triangle by using the coordinates of the vertices: \[ \vec{a} = (1 - 0)\hat{i} + (3 - 0)\hat{j} + (-1 - 0)\hat{k} = \hat{i} + 3\hat{j} - \hat{k} \] \[ \vec{b} = (2 - 0)\hat{i} + (1 - 0)\hat{j} + (1 - 0)\hat{k} = 2\hat{i} + \hat{j} + \hat{k} \]

Find the cross product of the vectors

The area of the triangle can be calculated using the magnitude of the cross product \( \vec{a} \times \vec{b} \). Calculate the cross product:\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & -1 \ 2 & 1 & 1 \end{vmatrix} \]Calculating the determinant gives:\[ \left(3(1) - (-1)(1)\right)\hat{i} - \left(1(1) - (-1)(2)\right)\hat{j} + \left(1(1) - 3(2)\right)\hat{k} \]\[ = 4\hat{i} + 3\hat{j} - 5\hat{k} \]

Calculate the magnitude of the cross product

To find the area of the triangle, calculate the magnitude of \( \vec{a} \times \vec{b} \):\[ |\vec{a} \times \vec{b}| = \sqrt{(4)^2 + (3)^2 + (-5)^2} \]\[ = \sqrt{16 + 9 + 25} = \sqrt{50} \]

Calculate the area of the triangle

The area of the triangle is half of the magnitude of the cross product:\[ \text{Area} = \frac{1}{2} \times \sqrt{50} = \frac{\sqrt{50}}{2} \approx 3.54 \] square units.

Key concepts

Cross Product

When you want to find the area of a triangle in 3D space, the cross product of two vectors plays a key role. In this particular context, vectors help describe the sides of the triangle. By calculating the cross product of these vectors, you determine a new vector that is perpendicular, or orthogonal, to the plane defined by the triangle. This is fundamentally useful because the magnitude (or length) of this cross product vector directly relates to the area of the parallelogram that the two vectors form.

And since a triangle is half of a parallelogram, you can straightforwardly use this relation to find the triangle's area. The cross product is mathematically represented as \( \vec{a} \times \vec{b} \), with a determinant that provides the basis for computation.

Determinant Calculation

Determinants are crucial when working with cross products, particularly in three dimensions. They provide a way to find the vector that results from the cross product of two other vectors. Using the components of vectors \(\vec{a}\) and \(\vec{b}\), the determinant is usually presented in a 3x3 matrix form. The first row contains unit vectors \(\hat{i}, \hat{j}, \hat{k}\), helping visualize and compute the cross product in a structured manner.

For our vectors: \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & -1 \ 2 & 1 & 1 \end{vmatrix}, \) we solve by expanding the determinant along the first row. Each component (involving \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \)) is obtained through a series of smaller 2x2 determinants, providing a vector result such as \(4\hat{i} + 3\hat{j} - 5\hat{k}\).

This result significantly aids further calculation, particularly in finding the magnitude necessary for determining the area of the triangle.

Vector Magnitude

Once you obtain the cross product vector, determining its magnitude is the essential next step to find the area of your triangle. The magnitude of a vector \( \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} \) is calculated using the formula: \( \sqrt{a^2 + b^2 + c^2} \). This formula measures the 'length' or 'size' of the vector in three-dimensional space.

For the previously calculated cross product vector from our problem, \(4\hat{i} + 3\hat{j} - 5\hat{k}\), the magnitude can be found as \(|\vec{v}| = \sqrt{4^2 + 3^2 + (-5)^2} = \sqrt{50}. \)

This magnitude is essential, not just a number, because when halved, it gives the area of the triangle formed by the original vectors, neatly tying together the concept of vector operations and geometric calculations.

Triangle Geometry

In 3D space, understanding the geometry of a triangle involves grasping how dimensions and vectors contribute to its structure. To determine the area, representation through vectors stemming from defined vertices is invaluable. Here, vectors like \(\vec{a}\) and \(\vec{b}\) start from one vertex and point to the other two vertices, effectively forming the edges of the triangle.

The geometric intuition is simple but profound: the vectors create a plane in 3D space, and their cross product offers a mathematical way to measure how they jointly cover that space. By leveraging half the magnitude of the cross product, it reveals the area of the triangle. This concrete relation between algebraic computations and spatial beliefs sums up the fundamental beauty of triangle geometry and vector mathematics.

It's a straightforward yet powerful method to transition from numbers and vectors to visual, spatial understanding.