Problem 13 - College Algebra and Calculus: An Applied Approach

Question

Find the distance between the point and the plane (see figure). The distance \(D\) between \(\mathrm{a}\) point \(\left(x_{0}, y_{0}, z_{0}\right)\) and the plane \(a x+b y+c z+d=0\) is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (0,0,0), 2 x+3 y+z=12 $$

Step-by-Step Solution

Answer

The distance D from the origin to the given plane is \(\frac{12\sqrt{14}}{14}\)

Step 1: Identify the given elements and equation

We are provided with a point at the origin of the coordinate system (0,0,0), and a plane defined by the equation \(2x+3y+z=12\). Therefore, the coefficients of \(x\), \(y\), and \(z\) in the equation are 2, 3, and 1 respectively. The constant term is -12. The formula for the distance (\(D\)) between a point \((x_0, y_0, z_0)\) and the plane \(ax+by+cz+d=0\) is given as \(D=\frac{ |ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}\)

Step 2: Substitute the parameter values in the distance equation

Substituting our parameter values into the formula yields \(D=\frac{ |2*0 + 3*0 + 1*0 -12|}{\sqrt{2^2 + 3^2 + 1^2}}\), which simplifies to \(D=\frac{ | -12|}{\sqrt{14}}\)

Step 3: Calculate the Distance

Solving the above equation gives \(D=\frac{12}{\sqrt{14}}\). Simplify the denominator to get the final answer in the simplest form: \(D=\frac{12\sqrt{14}}{14}\)

Key Concepts

Plane Equation

Understanding the equation of a plane is crucial when learning coordinate geometry. It is an algebraic expression that can be written in the standard form as (ax + by + cz + d = 0), where a, b, and c represent the plane's normal vector coefficients, and d represents the distance from the origin to the plane along this normal. A plane's normal vector is perpendicular to its surface, providing the directional components necessary to define the plane's orientation in three-dimensional space.

When provided with a plane equation such as (2x + 3y + z = 12), you can immediately know that the coefficients 2, 3, and 1 are the respective components of the normal vector, while 12 is the plane's displacement from the origin along that normal. The negative of this displacement is used when the plane's equation is rearranged to the standard form.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This method combines algebra and geometry to describe the position of points, lines, and shapes in a two- or three-dimensional space. For instance, a point in three-dimensional space is represented by its coordinates (x, y, z).

These coordinates correspond to the point's distances from the three mutually perpendicular axes, commonly referred to as the x-axis, y-axis, and z-axis. Understanding how to navigate this coordinate system is vital when determining various geometric figures' positions and properties, such as the distance from a point to a plane.

Algebraic Formula

An algebraic formula is an equation that allows you to solve mathematical problems with known and unknown variables. The distance formula used to find the distance between a point and a plane can be seen as an application of the absolute value concept and the Pythagorean theorem in three dimensions.

In practice, after identifying the plane's normal vector components (a, b, c) and the constant d, as well as the point's coordinates (x0, y0, z0), the algebraic formula |ax0 + by0 + cz0 + d| / sqrt(a^2 + b^2 + c^2) is applied. This formula computes the length of the orthogonal projection of the point onto the normal vector, yielding the shortest distance between the point and the plane.

Distance Calculation

Now, let's implement our understanding of the plane equation and coordinate geometry to calculate the distance between a point and a plane. Using the algebraic formula presented earlier, we substitute the known values from the plane's equation and the point's coordinates into the formula. For a point (0, 0, 0) and a plane defined by the equation (2x + 3y + z = 12), we follow the steps as described:

1. Substitute the values into the formula, obtaining: D = |2*0 + 3*0 + 1*0 - 12| / sqrt(2^2 + 3^2 + 1^2).
2. Simplify the equation to find D = | -12 | / sqrt(14), using the absolute value to ensure a non-negative distance.
3. To further simplify, we rationalize the denominator, resulting in the final distance D = 12 * sqrt(14) / 14.

This final expression represents the shortest distance between the point (0,0,0), and the plane, demonstrating a practical application of algebra in geometry.