Problem 30 - Differential Equations and Linear Algebra

Question

Use the result of Problem 28 to find the distance from the given point \(P\) to the given plane \(\mathcal{P}\). $$P(0,-1,3) ; \text { Plane } \mathcal{P} \text { with equation } 3 x-y-z=5$$

Step-by-Step Solution

Answer

The distance from point \(P(0, -1, 3)\) to plane \(\mathcal{P}\) with equation \(3x - y - z = 5\) is \(d = \frac{6}{\sqrt{11}}\).

Step 1: Identify coefficients of the plane's equation

Identify the values of \(A\), \(B\), \(C\), and \(D\) from the given equation of the plane: $$3x - y - z = 5$$ We have: $$A = 3$$ $$B = -1$$ $$C = -1$$ $$D = 5$$

Step 2: Identify coordinates of the given point

Identify the coordinates of point \(P(0, -1, 3)\): $$x_0 = 0$$ $$y_0 = -1$$ $$z_0 = 3$$

Step 3: Plug the values into the formula

Now, we are ready to plug all values into the formula: $$d = \frac{|3x_0 - y_0 - z_0 + 5|}{\sqrt{3^2 + (-1)^2 + (-1)^2}}$$

Step 4: Calculate the distance

Calculate the distance \(d\) using the values from Steps 1 and 2: $$d = \frac{|3(0) - (-1) - 3 + 5|}{\sqrt{3^2 + (-1)^2 + (-1)^2}}$$ $$d = \frac{|1 + 5|}{\sqrt{9 + 1 + 1}}$$ $$d = \frac{6}{\sqrt{11}}$$ So, the distance from point \(P\) to plane \(\mathcal{P}\) is: $$d = \frac{6}{\sqrt{11}}$$

Key Concepts

Plane Equation

Understanding plane equations is crucial for calculating distances from a point. A plane in three-dimensional space can be expressed by a linear equation of the form:

• \( Ax + By + Cz = D \)

These letters, \( A \), \( B \), and \( C \), are the coefficients of the variables and determine the plane's orientation.
The equation \( 3x - y - z = 5 \) specifies a plane in which the coefficient values are \( A = 3 \), \( B = -1 \), \( C = -1 \), and the constant term \( D = 5 \).
These coefficients not only define the plane's position but are essential for calculating the perpendicular distance from a point to this plane.

Distance Formula

The distance formula used to measure how far a point is from a plane is a specialized version in vector algebra. This formula is:

• \( d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \)

It requires both the plane equation coefficients \( A \), \( B \), \( C \), and the point coordinates.
You substitute these values into the formula to get the distance \( d \). The denominator is the magnitude of the normal vector to the plane, while the numerator is the absolute value of the plane equation with point coordinates plugged in.
This formula effectively calculates the perpendicular distance, thanks to vector algebra principles.

Coordinates

Coordinates define the location of points in space. Each point is represented by three numbers corresponding to the axes \( x \), \( y \), and \( z \).
For instance, the coordinates of point \( P \) are given as \( (0, -1, 3) \). Here:

• \( x = 0 \)
• \( y = -1 \)
• \( z = 3 \)

These coordinates are essential for calculating the distance to a plane, as they are substituted into the distance formula.
Knowing the precise location of a point is crucial for applications in geometry and vector algebra.

Vector Algebra

Vector algebra plays a fundamental role in calculating distances between geometrical objects. It helps unify various spatial concepts and operations.
A vector is defined by direction and magnitude; in this context, a normal vector to a plane is derived from the plane equation's coefficients.

• The normal vector is \( \langle A, B, C \rangle \).

This vector is vital because it helps define the plane's orientation and is used in the distance formula's denominator.
In essence, vector algebra processes such data to provide meaningful measurements, like the distance of a point from a plane, in a precise and efficient way.
Understanding these vectors is crucial for mastering spatial calculations in three-dimensional space.