Problem 21 - Precalculus Mathematics for Calculus

Question

Find an equation of the plane that passes through the points \(P, Q,\) and \(R\) $$P(6,-2,1), \quad Q(5,-3,-1), \quad R(7,0,0)$$

Step-by-Step Solution

Answer

Plane equation is \( x - y + z = 5 \).

Step 1: Identify Vectors

To find a plane equation using points, first find two vectors from the points given. Use vectors \( \vec{PQ} \) and \( \vec{PR} \). **Vector \( \vec{PQ} \):** \( Q - P = (5 - 6, -3 + 2, -1 - 1) = (-1, -1, -2) \). **Vector \( \vec{PR} \):** \( R - P = (7 - 6, 0 + 2, 0 - 1) = (1, 2, -1) \).

Step 2: Calculate Cross Product

The normal vector to the plane can be found using the cross product of vectors \( \vec{PQ} \) and \( \vec{PR} \). Calculate: \[ \vec{n} = \vec{PQ} \times \vec{PR} \].i\((1\cdot(-1) - (-1)\cdot2),\)\((2\cdot(-1) - (-1)\cdot1),\)\(((-1)\cdot(2) - (-1)\cdot1)\) = (1, -1, 1).

Step 3: Construct the Plane Equation

The general equation of a plane given a normal vector \( \vec{n} = (a, b, c) \) and a point \( P(x_0, y_0, z_0) \) is: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]. Substitute \( \vec{n} = (1, -1, 1) \) and point \( P(6, -2, 1) \): \[ 1(x - 6) - 1(y + 2) + 1(z - 1) = 0 \],yielding: \( x - y + z = 5 \).

Key Concepts

3D geometry

In 3D geometry, we're dealing with a world that adds depth to the familiar length and width. It's like moving from a flat sheet of paper to a box. Everything has an extra dimension, and we can visualize points, lines, and planes in this more realistic space.
A point in 3D space is defined by three coordinates:

• x (length)
• y (width)
• z (depth)

These coordinates give us a way to pinpoint exact locations in a three-dimensional space.
When we talk about planes, imagine slicing through a block of cheese - the flat surface of the cut represents a plane in 3D space. Just like how a line in 2D needs two points, a plane in 3D typically needs three points that aren't in a straight line to be defined.

vectors in space

Vectors are essential tools in 3D geometry as they handle both direction and magnitude. They're like arrows that tell us not only how far to travel but also where to go.
In 3D space, vectors are represented as combinations of three components:

• i - The x-component
• j - The y-component
• k - The z-component

For instance, a vector \(\vec{A} = (a_1, a_2, a_3)\) can guide us from one point to another by adding its components to the original point.
To illustrate, consider finding vectors between points on a plane, such as \( \vec{PQ} \) and \( \vec{PR} \). Calculate them by subtracting coordinates to find their respective components.
Vectors help define planes through cross products, further paving our understanding of the spatial relationships between different points and directions.

cross product

The cross product is an operation used specifically with vectors in three dimensions, delivering a vector that is perpendicular to the plane formed by the original vectors. This perpendicular vector is useful as it helps define orientation in space.
Here's how the cross product works:

• Take two vectors, \(\vec{PQ}\) and \(\vec{PR}\).
• Compute their cross product using the determinant method, recognizing the role of each component.

The result is a vector \(\vec{n}\) pointing directly away from the plane containing the initial vectors. For instance, if \(\vec{PQ} = (-1, -1, -2)\) and \(\vec{PR} = (1, 2, -1)\), then \[\vec{PQ} \times \vec{PR} = (1\cdot(-1)-(-1)\cdot2,\ \ 2\cdot(-1)-(-1)\cdot1,\ \ (-1)\cdot2-(-1)\cdot1) = (1, -1, 1)\]
This cross product provides the normal vector necessary to develop an equation of the plane in question.