Problem 18 - 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 \(P Q R\) . $$ P(-2,2,0), \quad Q(0,1,-1), \quad R(-1,2,-2) $$

Step-by-Step Solution

Answer

The area of the triangle is \( \frac{1}{2}\sqrt{6} \), and a unit vector perpendicular to plane \(PQR\) is \(\left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)\).

Step 1: Determine the Vectors

Given the points \( P(-2,2,0) \), \( Q(0,1,-1) \), and \( R(-1,2,-2) \), calculate the vectors \( \vec{PQ} \) and \( \vec{PR} \):\[ \vec{PQ} = Q - P = (0 - (-2), 1 - 2, -1 - 0) = (2, -1, -1) \]\[ \vec{PR} = R - P = (-1 - (-2), 2 - 2, -2 - 0) = (1, 0, -2) \]

Step 2: Find the Cross Product

Find the cross product of vectors \( \vec{PQ} \) and \( \vec{PR} \) to get a vector perpendicular to the plane of the triangle:\[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & -1 \ 1 & 0 & -2 \end{vmatrix} \]Calculating the determinant yields:\[ \vec{PQ} \times \vec{PR} = (-2 - 0)\hat{i} - (-2 + 1)\hat{j} + (0 + 1)\hat{k} = (-2, 1, 1) \]

Step 3: Calculate the Area of the Triangle

The area of the triangle is \( \frac{1}{2} \) the magnitude of the cross product found in Step 2:Calculate the magnitude:\[ |\vec{PQ} \times \vec{PR}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]Therefore, the area is:\[ \text{Area} = \frac{1}{2} \times \sqrt{6} \]

Step 4: Find the Unit Vector

To find a unit vector perpendicular to plane \( PQR \), divide the cross product found in Step 2 by its magnitude:Based on the cross product \((-2, 1, 1)\):\[ \text{Unit Vector} = \frac{1}{\sqrt{6}}(-2, 1, 1) = \left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) \]

Key Concepts

Cross Product

Understanding the cross product is key to vector calculus. It is a way to multiply two vectors in three-dimensional space to form a new vector that is perpendicular to the original two vectors. You can think of it as a tool that helps determine a direction orthogonal (at right angles) to a given plane.

In this case, we find the cross product between two vectors derived from points on the triangle's plane, namely vectors \( \vec{PQ} \) and \( \vec{PR} \). This can be calculated using the determinant of a matrix formed with the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) on the first row and the components of \( \vec{PQ} \) and \( \vec{PR} \) as subsequent rows:

\[\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & -1 \ 1 & 0 & -2 \end{vmatrix}\]

Calculating this determinant results in the vector \((-2, 1, 1)\). This vector is orthogonal to the plane formed by the points \( P, Q, \) and \( R \). The direction indicated by this vector gives us information about the orientation of the plane with respect to the axes in a three-dimensional space.

Unit Vector

A unit vector is crucial when you need a direction with a magnitude of one. It is often used to indicate direction without any scaling. Creating a unit vector from a given vector involves dividing each component of the vector by its magnitude. This process normalizes the vector.

In this exercise, we derived an orthogonal vector \((-2, 1, 1)\) using the cross product. To normalize, you calculate its magnitude:

• Magnitude: \( \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6} \)

By dividing each component of the vector by this magnitude, we achieve a unit vector perpendicular to the plane of triangle \( PQR \):

• \( \text{Unit Vector} = \frac{1}{\sqrt{6}}(-2, 1, 1) = \left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) \)

This unit vector maintains the direction given by the cross product while standardizing its length to one unit. This is useful for a consistent way to describe perpendicularity in space.

Area of a Triangle

Calculating the area of a triangle in three-dimensional space can be elegantly performed using the cross product. In this vector-based approach, you first determine two vectors along the edges of the triangle. The magnitude of their cross product offers a measure of the parallelogram spanned by these vectors.

The triangle's area is half that of the parallelogram, as it effectively divides this shape into two triangles:

• Area Formula: \( \text{Area} = \frac{1}{2} |\vec{PQ} \times \vec{PR}| \)

Previously, we determined the magnitude of the cross product \( \vec{PQ} \times \vec{PR} \) as \( \sqrt{6} \). Consequently, the area of triangle \( PQR \) is calculated as:

• \( \text{Area} = \frac{1}{2} \times \sqrt{6} \)

This vector-based method conveniently utilizes three-dimensional coordinates directly, avoiding the need for traditional trigonometry and equations. It demonstrates the power of vector calculus in solving geometrical problems in 3D space.