Problem 14 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (0,0,0), 8 x-4 y+z=8 $$

Step-by-Step Solution

Answer

The distance between the point and the plane is \(\frac{8}{9}\) units.

Step 1: Identify the given point and coefficients of the plane

In this problem, the given point is (0,0,0) and the equation of the plane is \(8x - 4y + z = 8\), from which we can determine that the coefficient 'a' is 8, 'b' is -4, 'c' is 1 and the constant 'd' is -8.

Step 2: Substitute these values into the distance formula

Using the formula \(D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\), we substitute our given values into the formula to obtain \(D=\frac{\left|8*0-4*0+1*0-8\right|}{\sqrt{8^{2}+(-4)^{2}+1^{2}}}\)

Step 3: Solve the expression

Now calculate both the numerator and the denominator. The numerator simplifies to 8. The denominator simplifies to \(\sqrt{64+16+1} = \sqrt{81}\). Thus the expression becomes \(D=\frac{8}{9}\).

Key Concepts

Point-Plane Distance Formula

In three-dimensional geometry, one of the fundamental tasks is to find out how far a given point is from a specific plane. This is crucial in fields like physics, engineering, and computer graphics. The distance between a point \((x_0, y_0, z_0)\)and a plane given by the equation \(ax + by + cz + d = 0\)is calculated using the Point-Plane Distance Formula. This formula is written as:

• \[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

The numerator of this formula \(|ax_0 + by_0 + cz_0 + d|\)represents the algebraic distance, a signed value that measures how far and in which direction the point is from the plane.
Meanwhile, the denominator\(\sqrt{a^2 + b^2 + c^2}\)calculates the magnitude of the plane's normal vector, which adjusts for directionality to ensure the distance is a non-negative number. Understanding this formula allows you to effectively compute the perpendicular distance from any point to any given plane.

3D Coordinate Geometry

Three-dimensional coordinate geometry, often referred to as 3D geometry, extends the principles of 2D geometry into an extra dimension. It involves three axes: the x-axis, y-axis, and z-axis, which together form the 3D coordinate system.
This system allows us to define locations of points in space using ordered triples \((x, y, z)\),where:

• \(x\) is the position along the horizontal axis,
• \(y\) is the position along the vertical axis,
• and \(z\) represents depth or height above or below a reference plane.

3D geometry is crucial because it helps us describe the real-world layout more accurately.
With its broad application, from calculating distances as in our current task to interpreting the position of celestial bodies, mastering 3D coordinate geometry enhances analytical and spatial reasoning skills that are integral in many scientific and engineering disciplines.

Distance Calculation Method

When calculating the distance between a point and a plane, it's essential to follow a clear and structured method to ensure accuracy. The steps involved are straightforward but important:

• First, identify the coordinates of the point, \((x_0, y_0, z_0)\), as well as the coefficients \(a, b, c,\) and the constant \(d\) from the plane equation \(ax + by + cz + d = 0.\)
• Next, substitute these values into the point-plane distance formula:\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
• Simplify the expression in the numerator by performing the arithmetic operations specified.
• Calculate the denominator, which is the square root of the sum of the squares of \(a, b,\) and \(c.\)
• Finally, divide the absolute value of the simplified numerator by the previously computed denominator to obtain the distance \(D.\)

This method ensures that the result represents the shortest path between the point and the plane, providing the exact perpendicular distance that is so often needed in various scientific analyses and real-world applications.