Problem 46 - University Calculus: Early Transcendentals

Question

Find the areas of the triangles whose vertices are given. $$A(0,0,0), \quad B(-1,1,-1), \quad C(3,0,3)$$

Step-by-Step Solution

Answer

The area of the triangle is \( \frac{3\sqrt{2}}{2} \).

Step 1: Calculate two vectors

Given the vertices of the triangle: \( A(0,0,0) \), \( B(-1,1,-1) \), and \( C(3,0,3) \). First, calculate two vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):\\[ \overrightarrow{AB} = B - A = (-1 - 0, 1 - 0, -1 - 0) = (-1, 1, -1) \] \[ \overrightarrow{AC} = C - A = (3 - 0, 0 - 0, 3 - 0) = (3, 0, 3) \]

Step 2: Find the cross product of vectors

Next, determine the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) to find a vector perpendicular to the plane containing \( AB \) and \( AC \):\[\overrightarrow{AB} \times \overrightarrow{AC} = \left| \begin{array}{ccc}i & \mathbf{j} & \mathbf{k} \ -1 & 1 & -1 \ 3 & 0 & 3 \end{array} \right| = \mathbf{i}(1 \times 3 - 0 \times (-1)) - \mathbf{j}(-1 \times 3 - (-1) \times 3) + \mathbf{k}(-1 \times 0 - 3 \times 1) \]Which simplifies to:\[= \mathbf{i}(3) - \mathbf{j}(-3 + 3) + \mathbf{k}(-3) = 3\mathbf{i} - 0\mathbf{j} - 3\mathbf{k} = (3, 0, -3)\]

Step 3: Calculate magnitude of the cross product

The magnitude of the cross product \( \| \overrightarrow{AB} \times \overrightarrow{AC} \| \) gives twice the area of the triangle. Calculate this magnitude:\[\sqrt{3^2 + 0^2 + (-3)^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2}\]

Step 4: Find the area of the triangle

The area \( A \) of the triangle is half the magnitude of the cross product:\[A = \frac{1}{2} \times 3\sqrt{2} = \frac{3\sqrt{2}}{2} \]

Key Concepts

Vectors in 3D Space

Vectors in 3D space are essential tools in geometry and physics. They allow us to describe objects or points using three dimensions: x, y, and z. Imagine vectors as arrows starting at a given point. Each arrow has a direction and length, which represents the vector's magnitude. For example, to find the vector \( \overrightarrow{AB} \) from point A to point B, we subtract A's coordinates from B's. This gives us \((-1, 1, -1)\). Similarly, the vector \( \overrightarrow{AC} \) is found by subtracting A's coordinates from C's, yielding \((3, 0, 3)\). By understanding these vectors, we gain insight into the spatial relationships between points and can further calculate areas, angles, and more. Vectors in 3D help us visualize complex arrangements in an understandable form.

Cross Product

The cross product is a fundamental operation that helps us find a vector perpendicular to two given vectors. If you have two vectors in space, their cross product results in another vector upright to both. It’s represented symbolically as \( \overrightarrow{AB} \times \overrightarrow{AC} \) in the given exercise. To calculate it, we use determinants from matrix algebra.
In the exercise, calculating the cross product gives us the vector \((3, 0, -3)\). This new vector is crucial because it tells us about the plane where vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) lie. Having this perpendicular vector grants us the ability to compute the area and understand more about the geometric arrangement of the triangle.

Magnitude of Vectors

The magnitude of a vector is its length - a numeric measure of how far the vector extends from its origin in space. To find a vector's magnitude, use the following formula: \( \sqrt{x^2 + y^2 + z^2} \). It's a straightforward application of the Pythagorean theorem in three dimensions. Consider the cross product vector from our exercise \((3, 0, -3)\), and let’s find its magnitude.
The result is \( \sqrt{3^2 + 0^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2} \). This magnitude has special significance because it represents twice the triangle's area spanned by the originating vectors. Understanding vector magnitudes aids us in quantifying dimensions, such as lengths and diagonals, in all sorts of geometric problems.

Geometry in Calculus

Geometry in calculus often bridges the precision of algebra with the intuitive aspects of geometry. In calculus, we study quantities with gradual change, which often involves geometric elements like the triangles within complex shapes. Calculating the area of such a triangle from a calculus standpoint requires breaking down the figures into simple parts and understanding key concepts such as the cross product.
Our exercise showed how the geometric structure of a triangle can be analyzed using vectors and their cross products, eventually linking to calculus-based area calculations. By considering operations like cross products and utilizing formula-driven techniques, we can solve geometric problems efficiently. Geometry in calculus provides us with powerful tools to model, analyze, and understand multidimensional shapes and surfaces.