Problem 18 - Precalculus Mathematics for Calculus

Question

Find a vector that is perpendicular to the plane passing through the three given points. $$P(3,4,5), Q(1,2,3), R(4,7,6)$$

Step-by-Step Solution

Answer

The vector \( (2, 0, -1) \) is perpendicular to the plane.

Step 1: Find two vectors in the plane

To find two vectors in the plane, consider vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). For \( \overrightarrow{PQ} \), subtract the coordinates of \( P \) from \( Q \): \( \overrightarrow{PQ} = (1-3, 2-4, 3-5) = (-2, -2, -2) \). Similarly, \( \overrightarrow{PR} \) is found by subtracting \( P \) from \( R \): \( \overrightarrow{PR} = (4-3, 7-4, 6-5) = (1, 3, 1) \).

Step 2: Calculate the cross product

The vector perpendicular to the plane is given by the cross product of \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). Calculate \( \overrightarrow{PQ} \times \overrightarrow{PR} \). For vectors \( \mathbf{a} = (-2, -2, -2) \) and \( \mathbf{b} = (1, 3, 1) \), the cross product is \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & -2 & -2 \ 1 & 3 & 1 \end{vmatrix} \).

Step 3: Compute the determinant

Evaluate the determinant to find the cross product: \( \mathbf{a} \times \mathbf{b} = \mathbf{i}((-2)(1) - (-2)(3)) - \mathbf{j}((-2)(1) - (-2)(1)) + \mathbf{k}((-2)(3) - (-2)(1)) \). Simplify the expression: \( \mathbf{i}(2 + 6) - \mathbf{j}(2 - 2) + \mathbf{k}(-6 + 2) = 8\mathbf{i} - 0\mathbf{j} - 4\mathbf{k} \).

Step 4: Resulting perpendicular vector

The resulting vector is \( (8, 0, -4) \). Divide by 4 to simplify: \( (2, 0, -1) \). This vector is perpendicular to the plane.

Key Concepts

Cross Product

The cross product is a fundamental operation used when dealing with vectors in 3D space. It results in a vector that is perpendicular to two given vectors, making it crucial in this exercise for finding a vector that is perpendicular to a plane.

When you start with two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), their cross product \( \mathbf{a} \times \mathbf{b} \) is given by the determinant formula:

• \( \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k} \)

In the exercise, we used the vectors \( \overrightarrow{PQ} = (-2, -2, -2) \) and \( \overrightarrow{PR} = (1, 3, 1) \). These vectors represent the sides of a triangle in the plane formed by the points \( P \), \( Q \), and \( R \). The cross product provides a vector \( \mathbf{8i} - \mathbf{4k} \) which is perpendicular to the plane through these points. This vector indicates a normal direction to the plane, a critical component when defining the plane's geometry.

Understanding cross products helps visualization in 3D geometry and is used in various applications, including physics.

Plane Equation

The plane equation provides a way to express a plane in 3D space using a point and a perpendicular vector (normal vector). By specifying a point on the plane and its normal vector, any plane can be described with a simple equation.

The standard form of a plane equation is:

• \( ax + by + cz = d \)

Where \( (a, b, c) \) is the normal vector found from the cross product. In this exercise, the normal vector is \( (2, 0, -1) \). Suppose we insert any point from the plane, such as \( P(3, 4, 5) \), into the equation. We calculate \( d \):

• \( 2(3) + 0(4) - 1(5) = 6 - 5 = 1 \)

So the equation is \( 2x + 0y - z = 1 \).

This expression is essential for various fields, as it allows us to handle complex spatial problems, map 3D spaces, and more. Knowing how to derive the plane equation is key for further studies in linear algebra and geometry.

Determinant

A determinant is a scalar value that is calculated from a square matrix. It provides essential information about the matrix, such as whether it is invertible or not, and is used heavily in vector calculations, including cross products.

For a 3x3 matrix, the determinant is calculated as follows:

• \( \begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \)

In the vector cross product \( \mathbf{a} \times \mathbf{b} \), we use the determinant of a 3x3 matrix made from the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of vectors \( \mathbf{a} \) and \( \mathbf{b} \).

Determinants are not just crucial for cross products but are also found in many other areas:

• Solving systems of linear equations
• Calculating volumes in higher dimensions
• Finding eigenvalues in linear algebra

By mastering determinants, students can tackle a wide array of mathematical challenges with increased confidence.