Problem 45 - Thomas Calculus

Question

Find the areas of the triangles whose vertices are given in Exercises \(41-47 .\) $$ A(1,0,0), \quad B(0,2,0), \quad C(0,0,-1) $$

Step-by-Step Solution

Answer

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

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

To find the area of a triangle given its vertices in 3D, we can use the formula: \[ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| \]where \(\vec{AB}\) and \(\vec{AC}\) are vectors in 3D formed from points A to B and A to C respectively, and \(\times\) denotes the cross product.

Step 2: Find Vectors AB and AC

Calculate the vectors \(\vec{AB}\) and \(\vec{AC}\) from the given vertices:- \(\vec{AB} = B - A = (0-1, 2-0, 0-0) = (-1, 2, 0)\)- \(\vec{AC} = C - A = (0-1, 0-0, -1-0) = (-1, 0, -1)\)

Step 3: Calculate the Cross Product \(\vec{AB} \times \vec{AC}\)

The cross product of \(\vec{AB} = (-1, 2, 0)\) and \(\vec{AC} = (-1, 0, -1)\) can be calculated as follows:\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 0 \ -1 & 0 & -1 \end{vmatrix} \]This results in: \[(2(-1) - 0(-1))\hat{i} - (0(-1) - 1(-1))\hat{j} + (-1(0) - (-1)(2))\hat{k} = (-2)\hat{i} - (1)\hat{j} + (2)\hat{k} \]\(\vec{AB} \times \vec{AC} = (-2, -1, 2)\).

Step 4: Compute the Magnitude of the Cross Product

The magnitude of the cross product \(\vec{AB} \times \vec{AC}\) is computed using the formula:\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{(-2)^2 + (-1)^2 + 2^2} \]Calculating this, we have:\[ \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Step 5: Calculate the Area of the Triangle

Finally, use the magnitude of the cross product to find the area of the triangle:\[ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} \times 3 = \frac{3}{2} \]

Key Concepts

Vector Cross Product

The vector cross product is a fundamental operation in 3D vector mathematics. It involves taking two vectors and producing a third vector that is perpendicular to both of them. To understand this, let's visualize three dimensions, the x, y, and z axes, forming a 3D space.

- **Definition**: Given two vectors, say \( \vec{u} = (u_1, u_2, u_3) \) and \( \vec{v} = (v_1, v_2, v_3) \), the cross product \( \vec{u} \times \vec{v} \) is computed using the determinant of a 3x3 matrix formed by unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and components of \( \vec{u} \) and \( \vec{v} \): \[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \] This expands to a vector \((c_1, c_2, c_3)\), where- - \(c_1 = u_2 v_3 - u_3 v_2\) - \(c_2 = -(u_1 v_3 - u_3 v_1)\) - \(c_3 = u_1 v_2 - u_2 v_1\)- **Properties**: The resulting vector is orthogonal to both \( \vec{u} \) and \( \vec{v} \), and its magnitude represents the area of the parallelogram that the vectors span.

Understanding the cross product is crucial for applications like determining the area of figures in 3D space, as it's pivotal in finding the perpendicular vector whose magnitude leads us to the desired area.

Magnitude of a Vector

To grasp the concept of the magnitude of a vector, think of it as the vector's length or size. This is an important measure when dealing with vectors in 3D geometry, as it allows us to quantify the extent of the vector.

- **Calculating Magnitude**: Suppose you have a vector \( \vec{v} = (v_1, v_2, v_3) \). The magnitude, denoted as \( \| \vec{v} \| \), is calculated using the formula: \[ \| \vec{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \] This formula is essentially an extension of the Pythagorean theorem to three dimensions.

- **Significance in Area Calculation**: In the context of finding the area of a triangle formed by three points in 3D space, the cross product of vectors derived from these points gives a vector whose magnitude represents the area of a parallelogram. The area of the triangle is then half of this magnitude.

So, by calculating the magnitude of the cross product, we derive the triangle's area. This is notably simpler than using direct geometric methods in 3D space and is foundational for many calculations in physics and engineering.

3D Geometry

3D geometry extends our understanding of shapes and spaces from flat surfaces to a world with depth. It becomes crucial when solving problems related to spaces and shapes that have three dimensions, like cubes, spheres, or triangles in a 3D space.

- **Basics of 3D Geometry**: In three-dimensional geometry, every point is represented by a set of three coordinates \((x, y, z)\). These coordinates tell us the location of points in a space that has width, height, and depth. This framework allows us to describe and analyze complex spatial relationships that are not possible in 2D.

- **Applications**: One of the practical applications of 3D geometry is in determining the area of shapes, such as triangles, positioned in a 3D space. This is done using vector mathematics to account for their orientation and size. For example, when given vertices of a triangle in space, you can find vectors between these points and use them to calculate cross products and magnitudes, as discussed.

- **Importance**: Understanding 3D geometry is vital in fields ranging from architecture, computer graphics to astronomy. As we explore the physical world or simulate virtual models, applying these concepts allows us to make accurate measurements and predictions.