Problem 13 - Differential Equations and Linear Algebra

Question

Find the distance from the point \(P(-1,3,5)\) to the plane with equation \(-x+3 y+3 z=8\).

Step-by-Step Solution

Answer

The distance from the point \(P(-1,3,5)\) to the plane with equation \(-x+3y+3z=8\) is \(\frac{17}{\sqrt{19}}\) units.

Step 1: Identify the coefficients and the constant term of the plane equation

We are given the equation of the plane: \(-x + 3y + 3z = 8\). Comparing it with the standard equation of a plane \(Ax + By + Cz = D\), we have: \[A = -1, B = 3, C = 3, \text{and } D = 8\]

Step 2: Identify the coordinates of the point P

The given point \(P(-1, 3, 5)\) has coordinates: \[x_0 = -1, y_0 = 3, \text{and } z_0 = 5\]

Step 3: Calculate the distance using the formula

Now, we will plug the values of A, B, C, x_0, y_0, z_0, and D into the distance formula: \[D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\] Substituting the values: \[D = \frac{|(-1)(-1) + (3)(3) + (3)(5) + 8|}{\sqrt{(-1)^2 + (3)^2 + (3)^2}}\] Evaluating the expression: \[D = \frac{|17|}{\sqrt{19}} = \frac{17}{\sqrt{19}}\] The distance from the point \(P(-1,3,5)\) to the plane with equation \(-x+3y+3z=8\) is \[\frac{17}{\sqrt{19}}\] units.

Key Concepts

Equation of a Plane

Understanding the equation of a plane is crucial when dealing with 3D geometry problems. The equation of a plane in three-dimensional space is typically written in the form \(Ax + By + Cz = D\). Here:

• \(A, B,\) and \(C\) are the coefficients that directly determine the orientation of the plane.
• \(x, y,\) and \(z\) represent the coordinates of any point on the plane.
• \(D\) is the constant term.

In the given exercise, the equation is \(-x + 3y + 3z = 8\). By comparing with the standard form, we identify:

• \(A = -1\)
• \(B = 3\)
• \(C = 3\)
• \(D = 8\)

This helps establish the plane in space and sets the foundation to calculate distances or understand its orientation.

Distance Formula

The distance formula is a mathematical approach used to determine the shortest distance between a point and a plane in three-dimensional space. For any given point \(P(x_0, y_0, z_0)\) and a plane defined by \(Ax + By + Cz = D\), the distance \(D\) from the point to the plane can be calculated using the formula:\[D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\]In simpler terms, the numerator \(|Ax_0 + By_0 + Cz_0 + D|\) represents the absolute value of the linear combination of the point's coordinates with the plane's coefficients and constant. The denominator \(\sqrt{A^2 + B^2 + C^2}\) normalizes this value by the vector's magnitude that is normal to the plane. Applying this to our specific problem with the point \(P(-1, 3, 5)\) and the plane \(-x + 3y + 3z = 8\), we substitute:\[D = \frac{|(-1)(-1) + (3)(3) + (3)(5) + 8|}{\sqrt{(-1)^2 + (3)^2 + (3)^2}}\]By evaluating the expression, we achieve the distance \(\frac{17}{\sqrt{19}}\), which gives us the shortest path length from point \(P\) to the plane.

Coordinate Geometry

Coordinate geometry provides a powerful framework for solving problems in various dimensions. It's all about positioning and relationships between points, lines, and planes in space. When dealing with solids in 3D space, we use the coordinate axes \(x, y,\) and \(z\) to establish a position for any object or point.In this context, knowing the coordinates of both points and planes can let us calculate distances, angles, and more. For instance, finding a point's distance to a plane helps understand positional relationships and is a core concept in fields like physics, computer graphics, and engineering.The core idea rests on calculus and linear algebra techniques to interpret and solve geometric problems. By mastering coordinate geometry, you can comfortably tackle 3D problem-solving, whether it's plotting a path, designing a structure, or analyzing spatial data.