Problem 30 - Essential Calculus Early Transcendentals

Question

\(21-32=\) Find an equation of the plane. The plane that passes through the points \((0,-2,5)\) and \((-1,3,1)\) and is perpendicular to the plane \(2 z=5 x+4 y\)

Step-by-Step Solution

Answer

The equation of the plane is \(-6x + 22y + 29z = 101\).

Step 1: Find the normal vector of the given plane

The given equation of the plane is \(2z = 5x + 4y\) which can be rewritten as \(5x + 4y - 2z = 0\). The normal vector to this plane is \(\langle 5, 4, -2 \rangle\).

Step 2: Find the direction vector of the line through the points

We identify the direction vector by subtracting coordinates of the two given points: \((0, -2, 5)\) and \((-1, 3, 1)\). The direction vector is \((-1 - 0, 3 + 2, 1 - 5) = \langle -1, 5, -4 \rangle\).

Step 3: Cross product to find the normal vector of the required plane

Taking the cross product of the normal vector \(\langle 5, 4, -2 \rangle\) with the direction vector \(\langle -1, 5, -4 \rangle\) to get the normal vector of the required plane.\[\begin{vmatrix}i \mathbf{i} & \mathbf{j} & \mathbf{k} \5 & 4 & -2 \-1 & 5 & -4 \end{vmatrix} = \langle (4)(-4) - (-2)(5), (-2)(-1) - (5)(-4), (5)(5) - 4(-1) \rangle = \langle -16 + 10, 2 + 20, 25 + 4 \rangle = \langle -6, 22, 29 \rangle\]

Step 4: Form the equation using the normal vector

To find the equation of the plane, use the point \((0, -2, 5)\) and the normal vector \(\langle -6, 22, 29 \rangle\). The equation is formed as follows:\[-6(x - 0) + 22(y + 2) + 29(z - 5) = 0\]

Step 5: Simplify the equation

Expanding and simplifying the equation: \[-6x + 22y + 44 + 29z - 145 = 0\]which simplifies further to \[-6x + 22y + 29z = 101\]. Thus, the equation of the plane is \[-6x + 22y + 29z = 101\].

Key Concepts

Normal Vector

A normal vector is a fundamental concept, especially in geometry and calculus. It is a vector that is perpendicular to a surface. In the context of a plane, the normal vector defines the orientation of the plane in space. To find the normal vector of a plane given by an equation like the one from our exercise, you can extract it directly from the coefficients of the variables. For example, given the plane equation in the form \(Ax + By + Cz = D\), the normal vector is \langle A, B, C \rangle\.

This normal vector will be perpendicular to any vector that lies on the plane. Knowing the normal vector is key to formulating the plane's equation and understanding how the plane is positioned in three-dimensional space.

Cross Product

The cross product is a mathematical operation that helps you find a vector perpendicular to two given vectors in three-dimensional space. It is central in the scenario where you need to find the normal vector of a plane formed by certain vectors or directions. The cross product of vectors \mathbf{a} = \langle a_1, a_2, a_3 \rangle\ and \mathbf{b} = \langle b_1, b_2, b_3 \rangle\ results in a third vector, \(abla \langle c_1, c_2, c_3 \rangle\)\, where:

• \c_1 = a_2b_3 - a_3b_2\
• \c_2 = a_3b_1 - a_1b_3\
• \c_3 = a_1b_2 - a_2b_1\

In the exercise provided, the cross product was used to find a new normal vector for the plane, given the normal vector of a previous plane and a newly found direction vector from a set of points. It’s these kind of computations that allow us to define orientations and angulations of elements in three-dimensional space.

Direction Vector

The direction vector represents the direction and magnitude between two points in space. It is found by taking the difference between the coordinates of these points. For instance, given two points \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\), the direction vector \(\mathbf{d}\) is calculated as \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle\.

This vector provides essential information when identifying characteristics of a line or plane, as seen in the exercise. The direction vector is often used to compute another vector, such as with the cross product, or to help define spaces like lines or planes.

Coordinates

Coordinates are numerical values that give exact positions for points in space. In three-dimensional space, coordinates are given as triples of the form \(x, y, z\). These values represent a point's position relative to three perpendicular axes: \x\ (horizontal), \y\ (vertical), and \z\ (depth).

Every point in three-dimensional geometry is defined by these values, and understanding how to use them helps in constructing geometric elements like lines and planes. For example, in this exercise, the given coordinates \(0, -2, 5\) and \(-1, 3, 1\) were used to determine direction vectors and ultimately assist in defining the plane's equation. Coordinates are foundational in plotting and visualizing geometric objects in space.