Problem 16 - Thomas Calculus

Question

In Exercises \(15-18\) , a. Find the area of the triangle determined by the points \(P, Q\) ,and \(R .\) b. Find a unit vector perpendicular to plane \(P Q R\) . $$ P(1,1,1), \quad Q(2,1,3), \quad R(3,-1,1) $$

Step-by-Step Solution

Answer

a. The area of the triangle is 3 square units. b. The unit vector perpendicular to the plane is \( \left( \frac{2}{3}, \frac{2}{3}, \frac{-1}{3} \right) \).

Step 1: Find Vectors PQ and PR

To find the area of the triangle, we start by finding vectors along two sides of the triangle. First, calculate vector \( \mathbf{PQ} \) by subtracting the coordinates of \( P \) from \( Q \): \( \mathbf{PQ} = Q - P = (2-1, 1-1, 3-1) = (1, 0, 2) \). Similarly, calculate vector \( \mathbf{PR} \) by subtracting the coordinates of \( P \) from \( R \): \( \mathbf{PR} = R - P = (3-1, -1-1, 1-1) = (2, -2, 0) \).

Step 2: Find the Cross Product

Next, find the cross product \( \mathbf{PQ} \times \mathbf{PR} \) to determine the vector perpendicular to both \( \mathbf{PQ} \) and \( \mathbf{PR} \). Using the formula for the cross product, \( \mathbf{PQ} \times \mathbf{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 2 \ 2 & -2 & 0 \end{vmatrix} = (4, 4, -2) \).

Step 3: Calculate the Magnitude of the Cross Product

The magnitude of the cross product gives twice the area of the triangle. Calculate the magnitude of the vector \( (4, 4, -2) \) by finding \( \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \).

Step 4: Find the Area of the Triangle

As the magnitude of the cross product gives twice the area of the triangle, divide by 2 to find the area: \( \text{Area} = \frac{6}{2} = 3 \).

Step 5: Find the Unit Vector Perpendicular to the Plane

The unit vector is found by dividing the cross product vector \( (4, 4, -2) \) by its magnitude. First, calculate the magnitude as \( 6 \) (from Step 3). Then, divide each component of the vector by 6: \( \left( \frac{4}{6}, \frac{4}{6}, \frac{-2}{6} \right) = \left( \frac{2}{3}, \frac{2}{3}, \frac{-1}{3} \right) \).

Key Concepts

Cross Product

The cross product is a mathematical operation that takes two vectors in three-dimensional space and produces another vector. This new vector is orthogonal or perpendicular to the original pair. To compute the cross product of two vectors, you use a special determinant-like structure:

• Arrange the vectors as rows in a matrix with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) at the top.
• Compute the determinant of this matrix, which results in a new vector.

In this particular problem, calculating the cross product \( \mathbf{PQ} \times \mathbf{PR} \) yields \( (4, 4, -2) \). This vector is crucial as it is perpendicular to both the original vectors \( \mathbf{PQ} \) and \( \mathbf{PR} \), making it useful for determining the perpendicular vector to a plane.
The cross product is especially useful in physics and engineering because it allows you to determine a normal vector to a plane in space, which is essential in various applications like torque and rotational dynamics.

Unit Vector

A unit vector is a vector with a magnitude of 1. It is often used in mathematics and physics to denote direction only, without any indication of magnitude. Finding a unit vector is a two-step process:

• First, compute the magnitude of the original vector.
• Then, divide each component of the vector by this magnitude.

In our exercise, after calculating the cross product which was \( (4, 4, -2) \), the next task was to find a unit vector. The magnitude of this cross product was calculated as 6. Each component of \( (4, 4, -2) \) is divided by 6 to get the unit vector \( \left( \frac{2}{3}, \frac{2}{3}, \frac{-1}{3} \right) \), ensuring that its magnitude is exactly 1.
Unit vectors are extremely useful in 3D geometry as they simplify problems that require direction normalization, like computer graphics and satellite orientation.

Area of Triangle

The area of a triangle in three-dimensional space can be efficiently calculated using the cross product. For triangles defined by points in 3D, the area is determined by:

• Forming vectors from the given points.
• Calculating the cross product of these vectors.
• Finding the magnitude of the resulting vector, which gives twice the area of the triangle.

Thus, dividing this magnitude by 2 gives the required area. In our problem, the magnitude of the cross product \( (4, 4, -2) \) was 6, resulting in a triangle area of \( \frac{6}{2} = 3 \).
This method is particularly efficient and widely used, as the cross product can easily compute the content of polygons in space by handling their respective vertices.

Perpendicular Vector

A perpendicular vector to a plane or line in 3D space is one that forms right angles with every vector that lies in that plane or along that line. In this exercise, the cross product \( \mathbf{PQ} \times \mathbf{PR} \) gives a perpendicular vector to the plane formed by points \( P, Q, \) and \( R \).
This perpendicular vector \( (4, 4, -2) \) is crucial in determining the orientation of the plane in three-dimensional space. The concept is employed in areas like computer graphics for normal vectors, which are used for lighting and shading effects.
Additionally, understanding perpendicular vectors is integral to solving problems involving planes and lines intersection, and checking if vectors are coplanar.

3D Geometry

3D geometry expands upon the principles of basic geometry by introducing depth as a dimension. In 3D spaces, vectors and points interact to form different shapes and planes.

• Coordinates represent each point with three numbers, typically \( x, y, \) and \( z \).
• Vectors like \( \mathbf{PQ} \) and \( \mathbf{PR} \) are defined by points and represent direction and magnitude.
• Planes consist of a collection of points extending infinitely in two dimensions.

Understanding the relationships between vectors, points, and planes is key to solving spatial problems involving distances, areas, and angles. In our problem, using vectors, we represent edges of a triangle and calculate both the area and a perpendicular vector, illustrating how 3D geometry assists in visualizing and solving complex spatial relationships often required in fields like architecture and engineering.