Problem 12 - Differential Equations and Linear Algebra

Question

Find the distance from the point \(P(0,0,0)\) to the plane with equation \(2 x-y+3 z=6\).

Step-by-Step Solution

Answer

The distance from the point \(P(0, 0, 0)\) to the plane with equation \(2x-y+3z=6\) is \(\frac{6}{\sqrt{14}}\).

Step 1: Find the normal vector

From the given equation of the plane, \(2x-y+3z=6\), we can directly read off the coefficients of the variables to identify the components of the normal vector. In this case, the normal vector \(\vec{n}\) would be: \[ \vec{n} = \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} \]

Step 2: Apply the point-to-plane distance formula

Now that we have the normal vector, we can use the point-to-plane distance formula to find the distance from the point \(P(0,0,0)\) to the plane. The formula is: \[ d = \frac{|ax + by + cz - d|}{\sqrt{a^2 + b^2 + c^2}} \] where \((x, y, z)\) are the coordinates of the point \(P\), \(a\), \(b\), and \(c\) are the components of the normal vector, and \(d\) is the constant term in the equation of the plane.

Step 3: Calculate the distance

We can now plug in the values to the formula and calculate the distance: \[ d = \frac{|2(0) - 1(0) + 3(0) - 6|}{\sqrt{2^2 + (-1)^2 + 3^2}} = \frac{|-6|}{\sqrt{14}} = \frac{6}{\sqrt{14}} \] The distance from the point \(P(0, 0, 0)\) to the given plane is \(\frac{6}{\sqrt{14}}\).

Key Concepts

Normal Vector

In geometry, a normal vector to a plane is a vector that is perpendicular to that plane. Understanding the concept of a normal vector is essential when dealing with the equations of planes.
For a plane described by the equation \(ax + by + cz = d\), the normal vector, denoted as \(\vec{n}\), can be directly obtained from the coefficients of \(x, y,\) and \(z\). This is because the equation of the plane can be interpreted as having a direction vector \(\vec{n} = \begin{pmatrix} a \ b \ c \end{pmatrix}\).

• In our problem, the equation of the plane is \(2x - y + 3z = 6\).
• The coefficients \(2, -1,\) and \(3\) give us the normal vector: \(\vec{n} = \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}\).

A normal vector acts as a directional guide for determining perpendicularity with respect to the plane. This is crucial when calculating distances from points to planes, as it helps maintain the integrity of the geometric relationships.

Point-to-Plane Distance Formula

The distance from a point to a plane is a fundamental concept in coordinate geometry. This distance is the shortest path from the point to the plane, running perpendicular to the plane. To find this distance, we use the point-to-plane distance formula.
The formula is written as:
\[d = \frac{|ax + by + cz - d|}{\sqrt{a^2 + b^2 + c^2}}\]

• \((x, y, z)\) are the coordinates of the point whose distance we are calculating.
• \(a, b,\) and \(c\) are the components of the normal vector \(\vec{n}\) to the plane.
• \(d\) is the constant on the right side of the plane's equation.

In our example:

• The point \(P(0,0,0)\) makes \(x = 0\), \(y = 0\), and \(z = 0\).
• The normal vector \(\vec{n}\) has components \(a = 2, b = -1,\) and \(c = 3\).
• Putting these into the formula, we get the distance \(d = \frac{|2(0) - (-1)(0) + 3(0) - 6|}{\sqrt{4 + 1 + 9}} = \frac{6}{\sqrt{14}}\).

This formula allows us to calculate the perpendicular distance efficiently, ensuring accuracy in various applications in geometry and calculus.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves the study of geometric figures using a coordinate system. Using this system, we can describe geometric shapes like points, lines, and planes using mathematical equations.

In our scenario, we have:

• A point described by coordinates \((0, 0, 0)\).
• A plane given by the equation \(2x - y + 3z = 6\).

By utilizing coordinate geometry, we derive exact geometric relationships and properties. One such important application is finding the shortest distance from a point to a plane, as demonstrated through the point-to-plane distance formula.
This discipline allows for precise intersection and distance calculations that would be more challenging using purely traditional geometric methods. It brings clarity and efficiency to solving complex geometric problems by leveraging algebraic formulas and methodologies.