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

The area of the triangle is 3, and the unit vector is \(\left( \frac{2}{3}, \frac{2}{3}, \frac{-1}{3} \right)\).

Step 1: Calculate vectors PQ and PR

To find the area of the triangle and a perpendicular vector to the plane, start by calculating the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\). Vector \(\overrightarrow{PQ}\) is found by subtracting the coordinates of \(P\) from \(Q\):\[\overrightarrow{PQ} = (2-1, 1-1, 3-1) = (1, 0, 2).\]Similarly, vector \(\overrightarrow{PR}\) is:\[\overrightarrow{PR} = (3-1, -1-1, 1-1) = (2, -2, 0).\]

Step 2: Find the cross product of PQ and PR

The cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\) gives a vector that is perpendicular to the plane of the triangle. Use the determinant formula for cross products:\[\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 0 & 2 \2 & -2 & 0 \\end{vmatrix}= (0-(-4), 4-0, -2-0) = (4, 4, -2).\]

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:\[||\overrightarrow{PQ} \times \overrightarrow{PR}|| = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6.\]

Step 4: Find the area of the triangle

Since the magnitude of the cross product is twice the area of the triangle, divide it by 2 to find the area:\[\text{Area} = \frac{1}{2} ||\overrightarrow{PQ} \times \overrightarrow{PR}|| = \frac{1}{2} \times 6 = 3.\]

Step 5: Determine the unit vector perpendicular to the plane

A unit vector is determined by dividing the vector by its magnitude. The perpendicular vector from the cross product is \((4, 4, -2)\). First find its magnitude:\[|| (4, 4, -2) || = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{36} = 6.\]Then, divide each component by the magnitude to find the unit vector:\[\left( \frac{4}{6}, \frac{4}{6}, \frac{-2}{6} \right) = \left( \frac{2}{3}, \frac{2}{3}, \frac{-1}{3} \right).\]

Key Concepts

Cross Product

The cross product of two vectors is a fundamental operation in vector calculus. It results in a new vector that is perpendicular to the plane formed by the two original vectors. Given vectors \( \overrightarrow{PQ} = (1, 0, 2) \) and \( \overrightarrow{PR} = (2, -2, 0) \), the cross product is calculated using the determinant method:

• Set up a matrix with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row.
• Place the components of \( \overrightarrow{PQ} \) in the second row.
• Place the components of \( \overrightarrow{PR} \) in the third row.

The resulting vector is \[(4, 4, -2)\], perpendicular to both \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). This orthogonal vector helps in finding properties like the area and provides direction for the plane created by these two vectors.

Unit Vector

A unit vector has a length (magnitude) of 1 and points in a particular direction. To find a unit vector that is perpendicular to the plane \(PQR\), we start with the previously calculated cross product vector \((4, 4, -2)\). Its magnitude is \(6\), calculated as follows:\[|| (4, 4, -2) || = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{36} = 6.\]A unit vector \((u)\) is obtained by dividing each component of the vector by its magnitude:\[\left( \frac{4}{6}, \frac{4}{6}, \frac{-2}{6} \right) = \left( \frac{2}{3}, \frac{2}{3}, \frac{-1}{3} \right).\]This unit vector is crucial because it defines a direction that is precisely perpendicular to the plane composed of the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). This means the vector "sticks out" from the plane, showing the normal direction.

Triangle Area

The area of a triangle in a 3-dimensional space determined by points \(P, Q,\) and \(R\) can be found via vector operations. Specifically, the cross product between vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \) gives a vector perpendicular to the plane of the triangle. The magnitude of this cross product vector gives twice the area of the triangle:\[||\overrightarrow{PQ} \times \overrightarrow{PR}|| = 6.\]To find the area of the triangle, divide the magnitude by 2:\[\text{Area} = \frac{1}{2} \times 6 = 3.\]This calculation reveals that vector calculus is not just about directions but also helps measure geometric properties like area. Seeing this in action demonstrates how multidimensional geometry is handled using algebraic operations.

Vector Operations

Vector operations, such as addition, subtraction, and cross products, are essential tools in vector calculus. Here, we start with subtraction to form vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \):

• \(\overrightarrow{PQ} = (1, 0, 2)\)
• \(\overrightarrow{PR} = (2, -2, 0)\)

These vectors are then used in the cross product operation to produce a vector perpendicular to the plane.Understanding vector operations enables one to navigate and compute complex geometrical properties. In this exercise, vectors formed the backbone of our calculations to find an area, calculate cross products, and ultimately derive perpendicular directions with unit vectors. By mastering these vector operations, you unravel the algebraic techniques that describe spatial relationships.