Problem 15 - University Calculus: Early Transcendentals

Question

a. Find the area of the triangle determined by the points \(P, Q\) and \(R\). b. Find a unit vector perpendicular to plane \(P Q R\). $$P(1,-1,2), \quad Q(2,0,-1), \quad R(0,2,1)$$

Step-by-Step Solution

Answer

a. Area = \(2\sqrt{30}\). b. Unit vector = \(\left(\frac{10}{\sqrt{120}}, \frac{2}{\sqrt{120}}, \frac{4}{\sqrt{120}}\right)\).

Step 1: Express Vectors from Points

To find the triangle's area in 3D space, first, express the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) from the given points. - For \(\overrightarrow{PQ}\): \((2-1, 0-(-1), -1-2) = (1, 1, -3)\)- For \(\overrightarrow{PR}\): \((0-1, 2-(-1), 1-2) = (-1, 3, -1)\)

Step 2: Calculate Cross Product

Calculate the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\). The cross product gives a vector perpendicular to both \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) and its magnitude helps find the triangle's area.- The cross product is calculated as follows:\[\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & -3 \ -1 & 3 & -1 \end{vmatrix}\]\[= \mathbf{i}(1 \cdot (-1) - (-3) \cdot 3) - \mathbf{j}(1 \cdot (-1) - (-3) \cdot (-1)) + \mathbf{k}(1 \cdot 3 - 1 \cdot (-1))\]\[= \mathbf{i}(1 + 9) - \mathbf{j}(1 - 3) + \mathbf{k}(3 + 1)\]\[= 10\mathbf{i} - (-2)\mathbf{j} + 4\mathbf{k}\]\[= 10\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\]

Step 3: Calculate Magnitude of Cross Product

The magnitude of \(\overrightarrow{PQ} \times \overrightarrow{PR}\) is calculated as:\[\|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \sqrt{10^2 + 2^2 + 4^2} = \sqrt{100 + 4 + 16} = \sqrt{120}\]

Step 4: Calculate Area of Triangle

The area of the triangle is half of the magnitude of the cross product:\[\text{Area} = \frac{1}{2} \times \sqrt{120} = \frac{\sqrt{120}}{2} = 2\sqrt{30}\]

Step 5: Find Unit Vector Perpendicular to Plane

The unit vector is found by normalizing the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\).- Normalize the vector \(10\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\):\[\text{Unit Vector} = \frac{1}{\sqrt{120}}(10\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\]- Simplify each component:\[= \left(\frac{10}{\sqrt{120}}, \frac{2}{\sqrt{120}}, \frac{4}{\sqrt{120}}\right)\]

Key Concepts

Vectors

Vectors are fundamental in 3D geometry, and they help to describe the direction and magnitude of an entity in space. Imagine a vector as an arrow pointing from one point to another. It has components along the x, y, and z axes, usually represented in the form \(a, b, c\). In our exercise, we define vectors that originate from the points \(P, Q,\) and \(R\).

• Direction: Indicated by the arrow's path, vectors determine the orientation in space.
• Magnitude: The vector's length, measuring the distance between its start and end points.

You can perform operations on vectors, such as addition, subtraction, and scalar multiplication, to derive new vectors or solutions to geometrical problems like the area of a triangle in 3D space.

Cross Product

The cross product is a crucial operation in vector mathematics, particularly in 3D geometry. It results in a vector that is perpendicular to two original vectors, making it fundamental in determining orientations in space.

• Calculating the Cross Product: Given two vectors, the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\) results in a vector perpendicular to both. The calculation employs determinants and specific vector components.
• Usefulness: The cross product is vital for finding normal vectors to planes, which helps in understanding angles and orientations.

In our example, the cross product allows us to find a vector pointing out of the plane formed by points \(P, Q,\) and \(R\). This is used later to find the area of the triangle and a perpendicular unit vector.

Magnitude

The magnitude of a vector is essentially its length in 3-dimensional space. Think of it as measuring how far a point is from the origin or another point in vector terms. Mathematically, the magnitude \(\|\overrightarrow{A}\|\) is determined using the Pythagorean theorem generalized in 3D:

• Formula: The magnitude of a vector \((a, b, c)\) is calculated as \sqrt{a^2 + b^2 + c^2}\.
• Application: In our exercise, finding the magnitude of the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\) provides the necessary length for further calculations, like determining the triangle's area.

Understanding how to calculate magnitude helps solve various spatial problems, revealing distances and effective comparisons between vector sizes.

Unit Vector

A unit vector is a vector with exactly one unit of length, used to describe direction without regard to magnitude. By normalizing vectors, you convert them into unit vectors.

• Normalization: To find a unit vector \(\overrightarrow{u}\) from a vector \(\overrightarrow{A}\), divide each component by its magnitude: \[\overrightarrow{u} = \frac{1}{\|\overrightarrow{A}\|}(a, b, c)\]
• Purpose: Unit vectors are crucial in defining directions in space, often acting as building blocks in vector mathematics and physics equations.

In our problem, the unit vector calculated from the cross product gives us a precisely oriented vector, perfectly perpendicular to the plane \(PQR\). This precise orientation is essential when studying complex spatial relationships and forces.