Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane through \((2,-3,2)\) and parallel to the plane of the vectors \(4 \mathbf{i}+3 \mathbf{j}-\mathbf{k}\) and \(2 \mathbf{i}-5 \mathbf{j}+6 \mathbf{k}\).

Step-by-Step Solution

Answer

The equation of the plane is \(13x - 26y - 26z = 52\).

Step 1: Determine the normal vector to the plane

The normal vector to the plane is perpendicular to both given vectors. To find the normal vector (\(\mathbf{n}\)), take the cross product of \(\mathbf{a} = 4\mathbf{i} + 3\mathbf{j} - \mathbf{k}\) and \(\mathbf{b} = 2\mathbf{i} - 5\mathbf{j} + 6\mathbf{k}\).

Step 2: Compute the cross product

Calculate \(\mathbf{a} \times \mathbf{b}\) using the determinant method.\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 3 & -1 \ 2 & -5 & 6 \end{vmatrix}= \mathbf{i}(3 \times 6 - (-5) \times (-1)) - \mathbf{j}(4 \times 6 - (-1) \times 2) + \mathbf{k}(4 \times -5 - 3 \times 2)\]Compute each part:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(18 - 5) - \mathbf{j}(24 - (-2)) + \mathbf{k}(-20 - 6)\]Hence,\[\mathbf{a} \times \mathbf{b} = 13\mathbf{i} - 26\mathbf{j} - 26\mathbf{k}\]

Step 3: Identify a point on the plane

We are given the point \((2, -3, 2)\) which lies on the plane.

Step 4: Write the equation of the plane

The general equation for a plane given a normal vector \((A, B, C)\) and a point \((x_0, y_0, z_0)\) on the plane is:\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]Substitute \(A = 13\), \(B = -26\), \(C = -26\), and \((x_0, y_0, z_0) = (2, -3, 2)\):\[ 13(x - 2) - 26(y + 3) - 26(z - 2) = 0 \]Expand this:\[ 13x - 26 - 26y - 78 - 26z + 52 = 0 \]

Step 5: Simplify the plane equation

Combine like terms:\[ 13x - 26y - 26z - 52 = 0 \] Reorder for the standard plane equation form:\[ 13x - 26y - 26z = 52 \]

Key Concepts

Cross Product

The cross product is a fundamental operation in vector mathematics that helps us find a vector that is perpendicular to two given vectors. This is particularly useful in three-dimensional space. For the exercise, the cross product is used to calculate a normal vector to the plane.The cross product of two vectors, \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), is given by the determinant of a matrix.Here’s how you can think about it:

• Place \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) at the top of the matrix.
• The second row contains the elements of \( \mathbf{a} \).
• The third row contains the elements of \( \mathbf{b} \).

You calculate the cross product \( \mathbf{a} \times \mathbf{b} \) as:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix}\]This formula results in a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), providing the necessary normal vector for the plane equation.

Normal Vector

The normal vector plays an essential role in determining the plane's orientation in space. In simple terms, a normal vector is perpendicular to the entire plane surface. This concept is crucial to formulating the plane equation.For this problem, we've calculated the normal vector by taking the cross product of the given vectors. Once you've found the normal vector, you can denote it as \( \mathbf{n} = A\mathbf{i} + B\mathbf{j} + C\mathbf{k} \).

• In this article, the normal vector was determined as \( 13\mathbf{i} - 26\mathbf{j} - 26\mathbf{k} \).
• The coefficients \( A, B, \) and \( C \) in the normal vector are used in the plane equation.

The normal vector ensures that the equation derived is applicable across the whole plane defined by this vector.

Point in Plane

A plane in space is determined by both a normal vector and a point in the plane. Identifying a specific point on the plane allows us to anchor the infinite possibilities of planes defined by the same normal vector to a specific location in space.For the problem at hand, the point \((2, -3, 2)\) was provided as lying on the plane. This guides the equation because it tells us precisely where the plane passes in the three-dimensional coordinate system.

• The point exists in coordinates \((x_0, y_0, z_0)\).
• It's used to substitute values into the general plane equation \( A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \).

The point ensures the plane equation is not only valid theoretically but also tethered to a real-world position.

Vectors

Vectors are essential mathematical objects used to represent quantities with both magnitude and direction. In this exercise, vectors help define the directionality and characteristics of the plane.Here, you started with two vectors, \( 4 \mathbf{i} + 3 \mathbf{j} - \mathbf{k} \) and \( 2 \mathbf{i} - 5 \mathbf{j} + 6 \mathbf{k} \), which are parallel to the plane we are examining. Vectors have applications in various fields like physics, engineering, and computer graphics.

• The direction each vector points to is determined by the coefficients of \( \mathbf{i}, \mathbf{j}, \text{and} \mathbf{k} \).
• Vectors can be added, scaled, and, as in this exercise, crossed to produce new vectors.

Understanding basic vector arithmetic and operations like the cross product is crucial for navigating through problems involving spatial dimensions.