Problem 32 - Thomas Calculus

Question

Find a plane through the points \(P_{1}(1,2,3), P_{2}(3,2,1)\) and perpendicular to the plane \(4 x-y+2 z=7\)

Step-by-Step Solution

Answer

The equation of the plane is \( 2x + 12y + 2z = 32 \).

Step 1: Find Normal Vector to the Given Plane

The given plane is described by the equation \(4x - y + 2z = 7\). The normal vector to this plane can be directly obtained from the coefficients of \(x\), \(y\), and \(z\). Thus, the normal vector is \(\mathbf{n} = (4, -1, 2)\).

Step 2: Find Direction Vector of the Line through Two Points

To determine the direction vector \(\mathbf{d}\) of the line passing through the points \(P_1(1,2,3)\) and \(P_2(3,2,1)\), subtract the coordinates of \(P_1\) from \(P_2\): \[ \mathbf{d} = (3-1, 2-2, 1-3) = (2, 0, -2). \]

Step 3: Find the Cross Product to Ensure Perpendicularity

Since we need a plane that passes through \(P_1\) and \(P_2\) and is perpendicular to the given plane, take the cross product of the normal vector \(\mathbf{n}\) and the direction vector \(\mathbf{d}\): \[ \mathbf{n} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & -1 & 2 \ 2 & 0 & -2 \end{vmatrix} = (2, 12, 2). \]This gives us a new normal vector for the desired plane.

Step 4: Formulate the Equation of the Plane

Using the newfound normal vector \( (2, 12, 2) \) and the point \( P_1(1,2,3) \), substitute into the plane equation format: \( 2(x - 1) + 12(y - 2) + 2(z - 3) = 0 \).Expanding and simplifying, we get:\[ 2x + 12y + 2z = 32. \]

Key Concepts

Plane Equation

A plane equation in analytical geometry is a way to represent a flat, two-dimensional surface in three-dimensional space. The general form of a plane equation can be written as \(Ax + By + Cz = D\), where \(A, B,\) and \(C\) are the coefficients that represent the components of the normal vector to the plane, and \(D\) is a constant. These coefficients are crucial because they define the orientation of the plane in space.
If you know a normal vector and a point through which the plane passes, you can easily formulate this equation. Just insert the point's coordinates into the formula \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\), where \((x_0, y_0, z_0)\) are the coordinates of the point on the plane. This form helps guide you to the standard \(Ax + By + Cz = D\) equation after expanding and simplifying. Understanding this equation is essential for tasks like finding intersections, calculating angles, and exploring distances in 3D space.

Cross Product

The cross product is a vector operation that takes two vectors in three-dimensional space and returns a third vector that is perpendicular to both of the original vectors. To compute the cross product, align the vectors such that they share the same initial point. This operation is particularly useful in finding a vector perpendicular to a given plane, which in turn can help define a new plane that meets specific conditions.

• The cross product of two vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\) is computed using a determinant of a matrix:

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1) \]The resulting vector is significant because it is orthogonal to both inputs, providing crucial vector orientations in spatial analyses.

Normal Vector

The normal vector to a plane is a vector that is perpendicular to the plane's surface. This vector is key to understanding and defining the orientation of planes in space. For example, in the plane equation \(Ax + By + Cz = D\), the vector \((A, B, C)\) is the normal vector.
Why is the normal vector useful? It helps in determining how the plane is positioned in three-dimensional space. If you have a point and a normal vector, you can easily find the plane equation. This vector can also be used to calculate angles between planes or determine if planes are parallel or perpendicular.
In practical applications, knowing the normal vector enables you to compute forces in physics problems, assess geometric alignments in architectural designs, and much more. Its role is fundamental in analytical geometry.

Direction Vector

A direction vector describes the orientation of a line within a plane or space. It is essentially a vector drawn from one point to another along the same line. In the context of the exercise, to find the direction vector \(\mathbf{d}\) between points \(P_1(1,2,3)\) and \(P_2(3,2,1)\), subtract the coordinates of the first point from the second: \(\mathbf{d} = (3-1, 2-2, 1-3) = (2, 0, -2)\).
This vector indicates the path one would travel from \(P_1\) to \(P_2\) on the line. Direction vectors allow us to express lines parametrically and are pivotal when assessing relations between lines and planes, such as finding intersections or establishing parallelism.

• They are instrumental in vector operations, like finding a line’s equation.
• They help in determining the magnitude and direction of the points in a plane.

The knowledge of direction vectors is a core aspect of vector calculus and plays a fundamental role in understanding the geometry of space.