Problem 45 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Formula for the Area of a Triangle Using Vertices

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

Step 2: Calculate Vectors AB and AC

To calculate \( \textbf{AB} \), subtract coordinates of \( A \) from \( B \):\[\textbf{AB} = B - A = (0 - 1, 2 - 0, 0 - 0) = (-1, 2, 0)\]To calculate \( \textbf{AC} \), subtract coordinates of \( A \) from \( C \):\[\textbf{AC} = C - A = (0 - 1, 0 - 0, -1 - 0) = (-1, 0, -1)\]

Step 3: Compute the Cross Product AB x AC

The cross product \( \textbf{AB} \times \textbf{AC} \) is given by:\[\textbf{AB} \times \textbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & 0 \ -1 & 0 & -1 \end{vmatrix}\]Evaluate the determinant:\[\mathbf{i}(2(-1) - 0) - \mathbf{j}( (-1)(-1) - 0 ) + \mathbf{k}( (-1)(0) - (-1)(2) )= -2\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}\]So, \( \textbf{AB} \times \textbf{AC} = (-2, -1, 2) \).

Step 4: Find the Magnitude of the Cross Product

The magnitude of a vector \( \textbf{v} = (x, y, z) \) is given by \( \sqrt{x^2 + y^2 + z^2} \). For our cross product:\[|\textbf{AB} \times \textbf{AC}| = \sqrt{(-2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\]

Step 5: Calculate the Area of the Triangle

Plug the magnitude of the cross product into the area formula:\[\text{Area} = \frac{1}{2} \times 3 = \frac{3}{2}\]

Step 6: Interpret the Result

The area of the triangle with vertices at \( A(1, 0, 0) \), \( B(0, 2, 0) \), and \( C(0, 0, -1) \) is \( \frac{3}{2} \).

Key Concepts

Cross Product

The cross product is an operation that is used with vectors, especially in three-dimensional geometry. It produces a new vector that is perpendicular to the plane formed by the initial vectors.

• The direction of the resultant vector follows the right-hand rule. If you align your right hand with the first vector and curl your fingers toward the second vector, your thumb points in the direction of the cross product.
• The cross product is denoted by a "×" symbol. If you have two vectors \( \mathbf{u} \) and \( \mathbf{v} \), their cross product is \( \mathbf{u} \times \mathbf{v} \).

A common application of the cross product is finding the area of parallelograms and triangles in 3D space. For instance, if you have two vectors \( \mathbf{AB} \) and \( \mathbf{AC} \), the cross product \( \mathbf{AB} \times \mathbf{AC} \) gives a vector perpendicular to the triangle's plane.Knowing how to compute it involves calculating the determinant of a matrix involving the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), and the components of the two vectors. This is shown in the step-by-step solution provided for the triangle from points \( A, B, \) and \( C \). The cross product \((-2, -1, 2)\) indicates we are dealing with a vector normal to the triangle.

Magnitude of a Vector

The magnitude of a vector is a measure of its length. Much like the length of a line segment, vectors have magnitudes that are straightforward to find.

• If you have a vector \( \mathbf{v} = (x, y, z) \), the magnitude is given by the formula \( \sqrt{x^2 + y^2 + z^2} \).
• To see it geometrically, the magnitude can be thought of as the distance from the origin (0,0,0) to the point (x,y,z) in three-dimensional space.

Magnitude is crucial when determining areas or volumes in geometry. After finding the cross product of two vectors, the magnitude tells us about the size of the area spanned by those initial vectors. As calculated, the magnitude of \((-2, -1, 2)\) is 3. This informs how large an area the vectors enclose when applied to the triangle.

Area of a Triangle

In 3D geometry, the area of a triangle can be deduced using the cross product of two vectors that define the triangle. This technique leverages the properties of the cross product and vector magnitude.

• Given vectors \( \mathbf{AB} \) and \( \mathbf{AC} \) calculated from the triangle's vertices, you find their cross product.
• The magnitude of this cross product vector gives twice the area of the triangle, because the resulting perpendicular vector relates to a parallelogram spanned by the two vectors.
• Therefore, dividing that magnitude by 2 yields the actual area of the triangle.

It provides a powerful method to calculate areas in three dimensions, offering exact results when verifying the size of spatial forms. In this case, the triangle formed by points \( A(1, 0, 0) \), \( B(0, 2, 0) \), and \( C(0, 0, -1) \) has an area of \( \frac{3}{2} \). This result is derived by taking half of the magnitude of the cross product, as triangles are half of the parallelogram area.