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

Step-by-Step Solution

Answer

a. The area is \( \frac{\sqrt{2}}{2} \). b. The unit vector is \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \).

Step 1: Find vectors PQ and PR

Calculate vector \( \overrightarrow{PQ} \) from point \( P \) to point \( Q \) by subtracting the coordinates: \[ \overrightarrow{PQ} = Q - P = (3 - 2, -1 + 2, 2 - 1) = (1, 1, 1) \]Similarly, calculate vector \( \overrightarrow{PR} \) from point \( P \) to point \( R \):\[ \overrightarrow{PR} = R - P = (3 - 2, -1 + 2, 1 - 1) = (1, 1, 0) \]

Step 2: Compute cross product of PQ and PR

Find the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) to get a vector perpendicular to the plane \(PQR\):\[\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 1 & 1 \1 & 1 & 0 \end{vmatrix}= \mathbf{i}(1 \cdot 0 - 1 \cdot 1) - \mathbf{j}(1 \cdot 0 - 1 \cdot 1) + \mathbf{k}(1 \cdot 1 - 1 \cdot 1)= -\mathbf{i} + \mathbf{j} + 0\mathbf{k}= (-1, 1, 0)\]

Step 3: Find the magnitude of the cross product

Calculate the magnitude of \( \overrightarrow{PQ} \times \overrightarrow{PR} \):\[ \left| (-1, 1, 0) \right| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2} \]

Step 4: Calculate the area of the triangle

The area \( A \) of triangle \(PQR\) is given by:\[ A = \frac{1}{2} \times \left| \overrightarrow{PQ} \times \overrightarrow{PR} \right| = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \]

Step 5: Calculate the unit vector perpendicular to the plane

To find the unit vector, divide the cross product by its magnitude:\[ \text{Unit Vector} = \frac{(-1, 1, 0)}{\sqrt{2}} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \]

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus. It is primarily used to find a vector that is perpendicular to two given vectors in three-dimensional space. Unlike the dot product, which results in a scalar, the cross product results in a vector.

To calculate the cross product of two vectors, we use the determinant of a matrix. For example, given vectors \( \overrightarrow{PQ} = (1, 1, 1) \) and \( \overrightarrow{PR} = (1, 1, 0) \) in the exercise, the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) is computed as follows:

• Write the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row.
• Write the components of \( \overrightarrow{PQ} \) in the second row: \( 1, 1, 1 \).
• Write the components of \( \overrightarrow{PR} \) in the third row: \( 1, 1, 0 \).

By expanding the determinant, we find the cross product is \( (-1, 1, 0) \).
This vector is perpendicular to both \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \), and hence perpendicular to the plane formed by these vectors.

Vectors

Vectors are essential in mathematics and physics, representing quantities having both magnitude and direction. In three-dimensional space, they are often represented as ordered triplets, such as \((x, y, z)\).

In the exercise, we calculate vectors using the points \(P(2, -2, 1)\), \(Q(3, -1, 2)\), and \(R(3, -1, 1)\). To find vector \( \overrightarrow{PQ} \), subtract the coordinates of \(P\) from \(Q\):

• \(Q_x - P_x = 3 - 2 = 1\)
• \(Q_y - P_y = -1 - (-2) = 1\)
• \(Q_z - P_z = 2 - 1 = 1\)

Consequently, \( \overrightarrow{PQ} = (1, 1, 1) \). Similarly, vector \( \overrightarrow{PR} \) is computed using the same method, resulting in \((1, 1, 0)\).
Vectors allow us to model and solve problems involving motion, forces, and other physical phenomena.

Unit Vector

A unit vector is a vector with a magnitude of one. It is used to indicate direction without concern for magnitude. To convert any vector into a unit vector, divide the vector by its magnitude.

In the exercise, we have the vector \((-1, 1, 0)\) from the cross product computation. To find the unit vector:

• First, compute the magnitude: \( \left| (-1, 1, 0) \right| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2} \).
• Then, divide each component of the vector by the magnitude: \((-1, 1, 0) / \sqrt{2} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)\).

This new vector, \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \), is a unit vector and points in the direction perpendicular to the plane defined by the triangle formed by points \(P\), \(Q\), and \(R\).
Unit vectors are extremely useful in physics for indicating directions, and in computer graphics for defining orientations.