Problem 26 - Algebra and Trigonometry

Question

Find an equation of the plane that passes through the points \(P, Q,\) and \(R .\) $$ P(2,0,0), \quad Q(0,2,-2), \quad R(0,0,4) $$

Step-by-Step Solution

Answer

The equation of the plane is \( 8x + 12y + 4z = 16 \).

Step 1: Find Two Vectors in the Plane

To find an equation of the plane, we need two vectors that lie in the plane. We can use the given points to find these vectors. Let's denote these points as follows: \( \vec{PQ} \) and \( \vec{PR} \).Calculate \( \vec{PQ} = Q - P = (0-2, 2-0, -2-0) = (-2, 2, -2) \).Calculate \( \vec{PR} = R - P = (0-2, 0-0, 4-0) = (-2, 0, 4) \).

Step 2: Find the Normal Vector to the Plane

The normal vector \( \vec{n} \) to the plane can be found by taking the cross product of the vectors \( \vec{PQ} \) and \( \vec{PR} \).Calculate \( \vec{n} = \vec{PQ} \times \vec{PR} \).\[\vec{n} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \-2 & 2 & -2 \-2 & 0 & 4 \\end{vmatrix}= \hat{i}(2 \cdot 4 - 0 \cdot -2) - \hat{j}(-2 \cdot 4 - (-2) \cdot -2) + \hat{k}(-2 \cdot 0 - 2 \cdot -2)\]\[= \hat{i}(8) - \hat{j}(-8 - 4) + \hat{k}(4)\]\[= (8, 12, 4)\]

Step 3: Form the Equation of the Plane

The equation of a plane can be derived using the point-normal form: \( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \), where \( (a, b, c) \) is a normal vector and \( (x_1, y_1, z_1) \) is a point on the plane.Using the normal vector \( \vec{n} = (8, 12, 4) \) and the point \( P(2, 0, 0) \), we have:\[8(x-2) + 12(y-0) + 4(z-0) = 0\]This simplifies to:\[ 8x + 12y + 4z = 16 \].

Key Concepts

Cross Product

The cross product is a crucial operation in vector algebra. It helps in finding a vector that is perpendicular to two given vectors.
This operation is beneficial when determining the normal vector of a plane, as it gives us a vector orthogonal to the plane.
Suppose you have two vectors \( \vec{A} = (a_1, a_2, a_3) \) and \( \vec{B} = (b_1, b_2, b_3) \).
The cross product \( \vec{A} \times \vec{B} \) is calculated as follows:

• \( \hat{i}(a_2b_3 - a_3b_2) \)
• \(-\hat{j}(a_1b_3 - a_3b_1) \)
• \( \hat{k}(a_1b_2 - a_2b_1) \)

These terms together form a new vector \((c_1, c_2, c_3)\).
The cross product helps find the normal vector in our plane equation problem by utilizing vectors \( \vec{PQ} \) and \( \vec{PR} \) found in the exercise.

Normal Vector

The normal vector is a key element in defining a plane in three-dimensional space.
This vector is orthogonal (perpendicular) to every vector that lies within the plane.
You can think of it as pointing straight out from the plane, like a flagpole from the ground.
In the equation of a plane, the coefficients of \(x\), \(y\), and \(z\) correspond to the components of a normal vector.
From our calculations, the normal vector to the plane formed by points \( P \), \( Q \), and \( R \) was found using the cross product, \( \vec{n} = (8, 12, 4) \).
This vector helps us describe the orientation of the plane in space.

Vector Operations

Vector operations include addition, subtraction, scaling, and the cross and dot products.
These operations allow us to manipulate vectors to understand their relationships better.
In this exercise, we used subtraction to find vectors \( \vec{PQ} \) and \( \vec{PR} \):

• Subtraction involves finding the difference between corresponding components.
• For example, \( \vec{PQ} = (0-2, 2-0, -2-0) = (-2, 2, -2) \).

These vector operations set the stage for finding the cross product, leading to the normal vector.
Understanding these basic vector operations is foundational for more complex vector calculations.

Point-Normal Form of a Plane

The point-normal form of a plane provides a straightforward method to define the plane's equation.
It is expressed as:

• \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \)

Here, \((a, b, c)\) are the components of a normal vector, and \((x_1, y_1, z_1)\) is a point on the plane.
By plugging the normal vector \( \vec{n} = (8, 12, 4) \) and the point \( P(2, 0, 0) \) into this form, we derived:

• \( 8(x - 2) + 12(y) + 4(z) = 0\)
• Simplifying gives us the equation: \( 8x + 12y + 4z = 16 \).

This equation precisely defines the plane containing points \( P \), \( Q \), and \( R \), utilizing both the normal vector and a specific point on the plane.