Problem 17 - 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 \(P Q R .\) $$ P(2,-2,1), \quad Q(3,-1,2), \quad R(3,-1,1) $$

Step-by-Step Solution

Answer

The area of the triangle is \( \frac{\sqrt{2}}{2} \) square units, and a unit vector perpendicular to the plane is \( (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0) \).

Step 1: Calculate the Vectors

To find the area of the triangle and the perpendicular unit vector, we first need the vectors formed by the points. We can find vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \) by subtracting the coordinates:\[ \overrightarrow{PQ} = Q - P = (3 - 2, -1 + 2, 2 - 1) = (1, 1, 1) \] \[ \overrightarrow{PR} = R - P = (3 - 2, -1 + 2, 1 - 1) = (1, 1, 0) \]

Step 2: Compute the Cross Product

The area of the triangle can be determined using the cross product of vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). The cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) is:\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(1\times0 - 1\times1) - \hat{j}(1\times0 - 1\times1) + \hat{k}(1\times1 - 1\times1) \]\[ = -\hat{i} + \hat{j} \] This simplifies to \( (-1, 1, 0) \).

Step 3: Calculate the Area of the Triangle

The magnitude of the cross product \(( -1, 1, 0)\) gives twice the area of the triangle. Calculate the magnitude:\[ \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \] So, the area of the triangle is:\[ \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \] square units.

Step 4: Find a Unit Vector Perpendicular to the Plane

The vector perpendicular to the plane is given by the cross product \((-1, 1, 0)\). To convert this to a unit vector, divide by its magnitude:\[ \text{Magnitude} = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2} \]Unit vector = \[ \left( \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right) = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right) \].

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus, especially when dealing with vectors in three-dimensional space. It results in a vector that is perpendicular to the plane formed by the initial two vectors. To compute the cross product of two vectors, we set up a determinant with the unit vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) in the top row and the components of the vectors in the following rows, like so:

• \( \overrightarrow{A} \times \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)

The result of this determinant is a new vector in three-dimensional space:

• \( (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k} \)

In practical terms, the cross product provides a vector that indicates the direction normal to a surface, which is helpful for various calculations including surface integrals and torque in physics. This operation is anti-commutative, meaning \( \overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A}) \).

Unit Vector

A unit vector is a vector with a magnitude of one. It maintains direction but without influence from magnitude, making it useful to describe directions in space. To convert any vector into a unit vector, we divide it by its magnitude.

• Given a vector \( \overrightarrow{v} = (x, y, z) \)
• The magnitude \(||\overrightarrow{v}|| = \sqrt{x^2 + y^2 + z^2} \)

The unit vector \( \overrightarrow{u} \) is then:

• \( \overrightarrow{u} = \frac{\overrightarrow{v}}{||\overrightarrow{v}||} = \left( \frac{x}{||\overrightarrow{v}||}, \frac{y}{||\overrightarrow{v}||}, \frac{z}{||\overrightarrow{v}||} \right) \)

In applications, unit vectors are crucial for indicating direction without altering magnitude, which is invaluable in geometry, physics, and engineering. The normalized direction assists in determining orientation in space without needing extensive calculations for magnitude.

Triangle Area

The area of a triangle can be efficiently calculated using vectors, particularly through the cross product method. When you have a triangle defined by points \(P\), \(Q\), and \(R\) in three-dimensional space, you can follow these steps:

• Identify vectors: \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\).
• Calculate the cross product: \(\overrightarrow{PQ} \times \overrightarrow{PR}\).
• The magnitude of the cross product gives you twice the area: \(||\overrightarrow{PQ} \times \overrightarrow{PR}||\).
• The actual area of the triangle is then: \(\frac{1}{2} ||\overrightarrow{PQ} \times \overrightarrow{PR}||\).

This method simplifies calculations in three-dimensional geometry, saving time and avoiding more complex traditional trigonometric methods. It’s especially helpful for irregular or complex shapes defined in a coordinate system.

Vectors in 3D Space

Vectors in three-dimensional space add complexity to our traditional understanding of lines and areas because they introduce a depth component. Each vector in 3D space is represented by three components, often denoting directions along the x, y, and z axes: \( (x, y, z) \).

• They extend from a point of origin and have both magnitude (length) and direction.
• Operations such as addition, subtraction, cross product, and dot product enable complex geometric transformations and calculations.

In practical applications:

• Addition might combine forces acting on an object.
• Subtraction can find the path between two points.
• Cross products find perpendicular directions and assist in calculating areas.
• Dot products help in finding angles between vectors or projecting one vector onto another.

Understanding vectors in 3D space is integral to fields like physics, engineering, and computer graphics, as they model real-world components realistically, aiding in simulations and design.