Problem 15 - Calculus Early Transcendentals

Question

Find the equation of the plane that is parallel to the vectors \langle 1,0,1\rangle and \(\langle 0,2,1\rangle,\) passing through the point (1,2,3)

Step-by-Step Solution

Answer

Answer: The equation of the plane is -2x - y + 2z - 2 = 0.

Step 1: Find the cross product of the parallel vectors

To find the normal vector \(\vec{n} = \langle A, B, C \rangle\), we will take the cross product of the given parallel vectors \(\vec{v_1} = \langle 1, 0, 1 \rangle\) and \(\vec{v_2} = \langle 0, 2, 1 \rangle\). The cross product of two vectors is given by: $\vec{n} = \vec{v_1} \times \vec{v_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right| $

Step 2: Evaluate the cross product

Evaluate the cross product determinant: $\vec{n} = \hat{i}(0-2) - \hat{j}(1-0) + \hat{k}(2-0) = -2\hat{i} - \hat{j} + 2\hat{k}$ We now have the normal vector \(\vec{n} = \langle -2, -1, 2 \rangle.\)

Step 3: Use the point-normal form of the plane equation

Now use the normal vector \(\vec{n} = \langle -2, -1, 2 \rangle\) and the given point \(P(1, 2, 3)\) to find the equation of the plane. The point-normal form of the plane equation is given by: \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\) Plug in the values, we get: \(-2(x-1) -1(y-2) + 2(z-3) = 0\)

Step 4: Simplify the equation

Now simplify the equation to find the standard form of the plane equation: \(-2x + 2 - y + 2 + 2z - 6 = 0\) \(-2x - y + 2z - 2 = 0\) Therefore, the equation of the plane that is parallel to the given vectors and passes through the point (1, 2, 3) is: \(\boxed{-2x - y + 2z - 2 = 0}\).

Key Concepts

Vector Cross Product

In vector mathematics, the cross product is an essential operation. It results in a new vector that is perpendicular to two original vectors in three-dimensional space. The cross product is often used to determine a plane's orientation or determine an object's rotation in physics.
To calculate the cross product, consider two vectors \( \vec{v_1} \) and \( \vec{v_2} \) defined by their components \( \langle a_1, b_1, c_1 \rangle \) and \( \langle a_2, b_2, c_2 \rangle \).
The formula for the cross product, \( \vec{v_1} \times \vec{v_2} \), is given by determining the determinant of a 3 x 3 matrix:

• \( \vec{n} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}\right| \)
• The resulting vector \( \vec{n} \) has components \( A, B, \) and \( C \):
• \( A = b_1c_2 - c_1b_2 \)
• \( B = c_1a_2 - a_1c_2 \)
• \( C = a_1b_2 - b_1a_2 \)

Understanding the cross product helps in finding vectors normal to surfaces, which aid in defining geometrical shapes and mathematical functions.

Normal Vector

A normal vector in a plane is perpendicular to the surface of the plane, playing a critical role in defining its orientation. In the process of finding the equation of a plane, a normal vector helps derive its formula and unique properties.
The calculation of the normal vector is often through operations like the cross product. For instance, let's say we have two vectors that define a plane. By taking the cross product of these vectors, you get a resultant vector that is perpendicular, hence the normal vector.
In our exercise, the vectors \( \langle 1, 0, 1 \rangle \) and \( \langle 0, 2, 1 \rangle \) are parallel to the plane. The cross product of these vectors \( \vec{n} = \langle -2, -1, 2 \rangle \) supplies us with the normal vector.
Applying the normal vector effectively assists in many calculations, including the formulation of equations to describe geometric forms such as planes.

Point-Normal Form

The point-normal form is a method used to describe a plane in a three-dimensional space using a normal vector and a specific point on the plane. This form offers a straightforward way to define the plane's equation.
Suppose you have a normal vector \( \vec{n} = \langle A, B, C \rangle \) and a point \( P(x_0, y_0, z_0) \). The point-normal form is used as follows:

• \( A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \)

Using the point-normal form provides clarity and ease when calculating equations of planes from specific geometrical conditions. In our scenario, inserting the normal vector \( \langle -2, -1, 2 \rangle \) and the point \( (1, 2, 3) \), we develop the equation \( -2(x-1) -1(y-2) + 2(z-3) = 0 \). This equation represents the plane passing through the point (1, 2, 3) and incorporating the normal vector for its orientation.

Simplifying Equations

Simplifying equations is the process of transforming complex mathematical expressions into simpler forms, enhancing comprehension and solving mathematical problems.
To simplify a plane's equation, gather like terms and perform arithmetic simplifications, as shown in the exercise walkthrough:

• Start with the initial equation: \( -2(x-1) -1(y-2) + 2(z-3) = 0 \)
• Distribute and collect like terms: \( -2x + 2 - y + 2 + 2z - 6 \)
• Simplify to: \( -2x - y + 2z - 2 = 0 \)

This process reduces the equation into a standard form, making it easier to interpret and verify. Simplification serves many mathematical purposes but predominantly allows easier manipulation for solving related problems or converting the methods into alternate mathematical languages.