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

Area: \( \frac{\sqrt{14}}{2} \); Unit Vector: \( \left(-\frac{2}{\sqrt{14}}, -\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\right) \).

Step 1: Determine the Vectors

First, compute the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \) using the given points. \( \overrightarrow{PQ} = Q - P = (0 - (-2), 1 - 2, -1 - 0) = (2, -1, -1) \) and \( \overrightarrow{PR} = R - P = (-1 - (-2), 2 - 2, -2 - 0) = (1, 0, -2) \).

Step 2: Find the Cross Product

To find a vector perpendicular to the plane, compute the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \). Using the determinant: \[ \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & -1 \ 1 & 0 & -2 \end{vmatrix} = \mathbf{i}(0 - 2) - \mathbf{j}(-4 - (-1)) + \mathbf{k}(0 - (-1)) = -2\mathbf{i} - 3\mathbf{j} + \mathbf{k}. \]

Step 3: Calculate Magnitude and Unit Vector

Compute the magnitude of the vector \( -2\mathbf{i} - 3\mathbf{j} + \mathbf{k} \):\[\sqrt{(-2)^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}.\]The unit vector is:\[\left(-\frac{2}{\sqrt{14}}, -\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\right).\]

Step 4: Area of Triangle Using Cross Product Magnitude

The area of the triangle \( \triangle PQR \) is given by half the magnitude of the cross product of \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \), which we computed in Step 2. Thus, the area is:\[\frac{1}{2} \times \sqrt{14}.\]

Key Concepts

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors in three-dimensional space. It's particularly handy in problems involving planes, like when you need a normal vector. The cross product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is calculated using the determinant format with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row:

• \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)

This results in a new vector \((a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\).
It's essential in physics and engineering because it can be used to compute torques and rotational forces.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length, is found using the Pythagorean theorem in three dimensions. For a vector \( \mathbf{v} = (x, y, z) \), the magnitude is calculated as \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \).
This is a crucial aspect in many applications, including physics, where it might represent speed or force.
Knowing the magnitude helps in normalizing the vector to find a unit vector or simply to understand the size and scale of the vector in 3D space.

Unit Vector

A unit vector is a vector with a magnitude of one. It keeps the same direction as the original vector but rescales it so that its length is one.
To find a unit vector, divide each component of the vector by the vector's magnitude.
For a given vector \( \mathbf{v} = (x, y, z) \), the unit vector \( \mathbf{u} \) is \( \mathbf{u} = \left( \frac{x}{\| \mathbf{v} \|}, \frac{y}{\| \mathbf{v} \|}, \frac{z}{\| \mathbf{v} \|} \right) \).
Unit vectors are particularly useful in defining directions, which is helpful in many scientific fields like navigation and computer graphics.

Area of a Triangle in 3D

The area of a triangle in three-dimensional space can be found using the cross product of two sides of the triangle.
Given three points \( P, Q, R \), you can form two vectors, such as \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \).
Once the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) is found, the magnitude of this vector gives twice the area of the triangle. Thus, the area \( A \) is computed as:

• \( A = \frac{1}{2} \times \text{magnitude of the cross product} \)

This formula helps in scenarios such as calculating land areas in surveying or finding the surface area involved in physical problems.

Perpendicular Vector to a Plane

Finding a vector that is perpendicular to a plane involves taking the cross product of two non-parallel vectors that lie in that plane.
In other words, for two vectors \( \mathbf{a} \) and \( \mathbf{b} \) in a plane, \( \mathbf{a} \times \mathbf{b} \) gives a vector normal to the plane.
Such a perpendicular vector is crucial in determining rotational axes or in physics when assessing forces acting at angles. It also defines the orientation of the plane in space, which can be useful in engineering and architecture.