Problem 31 - 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 \(\mathcal{P}\). \(P(-1,1,-1) ;\) Plane \(\mathcal{P}\) with equation \(z=2 x\)

Step-by-Step Solution

Answer

The distance from point \(P(-1, 1, -1)\) to the plane \(\mathcal{P}: -2x + 0y + z = 0\) is \(\frac{1}{\sqrt{5}}\).

Step 1: Rewrite the equation of the plane in general form

We are given that plane \(\mathcal{P}\) has the equation \(z = 2x\). We can rewrite this in the general form \(Ax + By + Cz = D\), by moving the "2x" to the left-hand side of the equation and adding an "0y" term. The rewritten equation of the plane is \(-2x + 0y + z = 0\).

Step 2: Identify A, B, C, D, and coordinates of P

From the equation \(-2x + 0y + z = 0\), we can identify the values of A, B, C, and D: \(A = -2, B = 0, C = 1, D = 0\) Additionally, we are given the coordinates of point P: \(P(-1, 1, -1)\)

Step 3: Apply the distance formula

The formula for the perpendicular distance from a point to a plane is: Distance \(=\frac{|Ax_0+By_0+Cz_0-D|}{\sqrt{A^2+B^2+C^2}}\) Using the identified values of A, B, C, D, and the coordinates of point P, we have: Distance \(=\frac{|(-2)(-1) + (0)(1) + (1)(-1) - 0|}{\sqrt{(-2)^2 + (0)^2 + (1)^2}}\)

Step 4: Simplify and calculate the distance

Now we just need to simplify and calculate the distance: Distance \(=\frac{|2 - 1|}{\sqrt{4 + 0 + 1}}\) Distance \(=\frac{1}{\sqrt{5}}\) Thus, the distance from point P to plane \(\mathcal{P}\) is \(\frac{1}{\sqrt{5}}\).

Key Concepts

Plane Equation in General Form

Understanding the equation of a plane in its general form is crucial when solving coordinate geometry problems. Essentially, this form is an algebraic expression that represents all the points that lie on the plane. The general form equation is typically given as \(Ax + By + Cz = D\), where \(A\), \(B\), and \(C\) are the coefficients that create a normal vector to the plane, and \(D\) represents the plane's distance from the origin, scaled by the normal vector.

Importantly, each of the variables \(x\), \(y\), and \(z\) in the equation corresponds to the respective coordinates of any point on the plane. If you're given a plane equation in a different format, such as \(z = 2x\), you'll want to rearrange it into the general form. This involves moving terms to one side of the equation to get a zero on the other side, and ensuring all three variable terms are present, even if it means adding a \(0y\) term, as seen in the example of converting \(z = 2x\) to \(-2x + 0y + z = 0\).

Perpendicular Distance Formula

The formula for calculating the perpendicular distance from a point to a plane is a vital tool in coordinate geometry. It takes the form \( \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}} \), where \((x_0, y_0, z_0)\) are the coordinates of the point and \(A, B, C,\) and \(D\) are parameters derived from the plane's general form equation.

The numerator of this fraction is the absolute value of the dot product of the plane's normal vector and a vector composed from the point's coordinates, offset by the constant \(D\). The denominator is the magnitude of the normal vector to the plane. This formula directly applies the concept that the shortest distance from a point to a plane is along the line perpendicular to it. Thus, this calculation yields the length of this perpendicular segment.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that involves the study of geometric figures through the use of coordinates and the principles of algebra and analysis. It serves as the foundation for many geometric problems, including those involving distances between points, lines, and planes.

The power of coordinate geometry lies in its ability to translate geometric problems into algebraic ones, which can then be solved using equations. For instance, to find the distance from a point to a plane, you can use algebraic methods to apply the perpendicular distance formula. This formula, as explained earlier, takes the geometric notion of a minimum distance and translates it into a solvable algebraic expression.

Coordinate geometry is widely used in various fields such as physics, engineering, computer science, and is integral to the study of calculus and higher mathematics.