Problem 29 - Essential Calculus Early Transcendentals

Question

(a) Find a nonzero vector orthogonal to the plane through the points \(P, Q,\) and \(R,\) and (b) find the area of triangle \(P Q R .\) $$P(1,0,1), \quad Q(-2,1,3), \quad R(4,2,5)$$

Step-by-Step Solution

Answer

Orthogonal vector: (0, 18, -9); Area: \(\frac{\sqrt{405}}{2}\).

Step 1: Find vectors PQ and PR

First, we calculate the vectors \( \vec{PQ} \) and \( \vec{PR} \) using the coordinates of points \( P(1,0,1), \, Q(-2,1,3), \, \text{and} \, R(4,2,5) \). The vector \( \vec{PQ} \) is found by subtracting the coordinates of \( P \) from \( Q \): \[ \vec{PQ} = (-2 - 1, \, 1 - 0, \, 3 - 1) = (-3, 1, 2) \] Similarly, the vector \( \vec{PR} \) is: \[ \vec{PR} = (4 - 1, \, 2 - 0, \, 5 - 1) = (3, 2, 4) \]

Step 2: Calculate the cross product of PQ and PR

To find the normal vector to the plane, compute the cross product \( \vec{PQ} \times \vec{PR} \). This cross product will yield a vector orthogonal to the plane:\[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 1 & 2 \ 3 & 2 & 4 \end{vmatrix} \] The determinant is computed as:\[ \vec{PQ} \times \vec{PR} = \mathbf{i}(1\cdot4 - 2\cdot2) - \mathbf{j}(-3\cdot4 - 2\cdot3) + \mathbf{k}(-3\cdot2 - 1\cdot3) \] \[ = \mathbf{i}(4-4) - \mathbf{j}(-12-6) + \mathbf{k}(-6-3) \] \[ = \mathbf{i}(0) + \mathbf{j}(18) + \mathbf{k}(-9) \] \[ = (0, 18, -9) \]

Step 3: Find the magnitude of the cross product

To find the area of the triangle \( \triangle PQR \), calculate the magnitude of the cross product vector, which is \((0, 18, -9)\). The magnitude is given by:\[ \| \vec{PQ} \times \vec{PR} \| = \sqrt{0^2 + 18^2 + (-9)^2} \]\[ = \sqrt{0 + 324 + 81} \]\[ = \sqrt{405} \]

Step 4: Calculate the area of triangle PQR

The area of the triangle is half the magnitude of the cross product calculated in the previous step:\[ \text{Area of } \triangle PQR = \frac{1}{2} \| \vec{PQ} \times \vec{PR} \| = \frac{1}{2} \times \sqrt{405} \]\[ = \frac{\sqrt{405}}{2} \]

Step 5: Conclusion

The nonzero vector orthogonal to the plane through points \( P, Q, R \) is \((0, 18, -9)\). The area of the triangle \( PQR \) is \( \frac{\sqrt{405}}{2} \).

Key Concepts

Cross Product

The cross product is a way to multiply two vectors that results in a new vector. This new vector is orthogonal, or perpendicular, to the original two vectors. The process of finding the cross product involves a determinant calculation with unit vectors \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \), which are the basis vectors for three-dimensional space.

• Given two vectors \( \vec{A} = (a_1, a_2, a_3) \) and \( \vec{B} = (b_1, b_2, b_3) \), the cross product \( \vec{A} \times \vec{B} \) is calculated with the determinant:

\[\vec{A} \times \vec{B} = \mathbf{i} (a_2 b_3 - a_3 b_2) - \mathbf{j} (a_1 b_3 - a_3 b_1) + \mathbf{k} (a_1 b_2 - a_2 b_1)\]
In this exercise, the vectors \( \vec{PQ} = (-3, 1, 2) \) and \( \vec{PR} = (3, 2, 4) \) were used. The result, \( (0, 18, -9) \), is a vector orthogonal to the plane containing the points \( P, Q, \, \text{and} \, R \).
This orthogonal vector can also be viewed as a normal vector, which is key in geometrical interpretations of planes.

Orthogonal Vectors

In vector calculus, two vectors are orthogonal if their dot product is zero. Orthogonal vectors are perpendicular to each other. This concept is important in many areas, including physics and engineering, because a single vector can represent diverse directions in three-dimensional space.

• If \( \vec{A} \cdot \vec{B} = 0 \), then \( \vec{A} \) and \( \vec{B} \) are orthogonal.
• The vector \( (0, 18, -9) \) is orthogonal to the plane formed by vectors \( \vec{PQ} \) and \( \vec{PR} \).

The orthogonal vector in a plane problem provides a direction that is unique to the arrangement of the initial vectors. It helps to define the plane in three-dimensional space and provides information about its orientation.

Triangle Area

Determining the area of a triangle in a three-dimensional space often involves vectors. The area is found using the cross product of two vectors that lie along two sides of the triangle. The magnitude of the cross product gives twice the area of the triangle.

• The formula for the area is:
  \[ \text{Area} = \frac{1}{2} \| \vec{A} \times \vec{B} \|\]
• The magnitude of the cross product \( (0, 18, -9) \) is \( \sqrt{405} \).
• Therefore, the area of triangle \( PQR \) is \( \frac{\sqrt{405}}{2} \).

This method avoids complex trigonometric calculations and directly uses vector properties to simplify the task, making it especially useful in higher dimensions or less straightforward geometric configurations.

Magnitude of a Vector

The magnitude of a vector represents its length. For a vector \( \vec{v} = (x, y, z) \), the magnitude \( \| \vec{v} \| \) is calculated as the square root of the sum of the squares of its components:
\[ \| \vec{v} \| = \sqrt{x^2 + y^2 + z^2} \]
Finding the magnitude is essential in vector calculus to determine distances and lengths, such as the area calculation where the cross product's magnitude is used.

• For the cross product vector \( (0, 18, -9) \), the magnitude is \( \sqrt{0^2 + 18^2 + (-9)^2} = \sqrt{405} \).
• This value was used to calculate the area of triangle \( PQR \).

Understanding how to find the magnitude of a vector helps in grasping other vector operations and applications, such as normalizing vectors or finding direction cosines.