Problem 18 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (2,-1,0), 3 x+3 y+2 z=6 $$

Step-by-Step Solution

Answer

The distance from the point (2,-1,0) to the plane 3x + 3y + 2z = 6 is \(\frac{3\sqrt{22}}{22}\).

Step 1: Identify the given variables

The given point coordinates are \((2,-1,0)\) where \(x_{0} = 2\), \(y_{0} = -1\), and \(z_{0} = 0\). The given plane equation is \(3x + 3y + 2z = 6\) where \(a = 3\), \(b = 3\), \(c = 2\), and \(d = -6\).

Step 2: Substitute the given variables into the distance formula

Plugging in the values into the distance formula: \(D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\), yields: \(D = \frac{|3*2 + 3*(-1) + 2*0 + (-6)|}{\sqrt{3^2 + 3^2 + 2^2}}\).

Step 3: Simplify the equation

Solving the equation yields: \(D = \frac{|6 -3 -6|}{\sqrt{9+9+4}} = \frac{|-3|}{\sqrt{22}} = \frac{3}{\sqrt{22}}\).

Step 4: Rationalize the denominator

To remove the square root in the denominator, you should multiply the numerator and the denominator by \(\sqrt{22}\). This gives: \(D = \frac{3\sqrt{22}}{22}\).

Key Concepts

Plane Equation

Understanding the equation of a plane is crucial when dealing with three-dimensional space in geometry. A plane is a flat, two-dimensional surface that extends infinitely in all directions. Its equation can be written in the form
\( ax + by + cz + d = 0 \),
where \(a\), \(b\), and \(c\) are coefficients that determine the orientation of the plane in three-dimensional space, and \(d\) is the scalar that determines the plane's position relative to the origin. Algebraically, this equation represents all the points \((x, y, z)\) that lie on a particular plane.

In the context of our exercise, the given plane equation is \(3x + 3y + 2z = 6\). Here, we can see that \(a=3\), \(b=3\), \(c=2\), and after rearranging \(d=-6\), providing us with the necessary values to understand the plane's orientation and position in space.

Distance Formula

The distance formula is a key concept in geometry that allows us to calculate the shortest distance between a point and a plane in three-dimensional space. It is expressed algebraically as
\( D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \),
where \(D\) represents the distance from the point \((x_0, y_0, z_0)\) to the plane \(ax + by + cz + d = 0\). The numerator involves the absolute value of the plane's equation substituted with the point's coordinates, which calculates the perpendicular distance to the plane. In the denominator, the square root of the sum of the squares of the plane's coefficients represents the normalization of the plane's normal vector, which is important to ensure the distance is calculated perpendicularly.

Algebraic Operations

Algebraic operations form the backbone of solving equations in mathematics. They include addition, subtraction, multiplication, and division, as well as more complex operations like factoring, expanding, and simplifying expressions. To solve for distance in our exercise, these basic operations are used systematically.

For instance, plugging the given point \((2,-1,0)\) into the distance formula requires multiplication and addition: \(3*2\) for the x-coordinate, \(3*(-1)\) for the y-coordinate, and \(2*0\) for the z-coordinate. These values are then summed, taking care to include the plane's scalar \(d\). The square root operation is involved in finding the magnitude of the plane's normal vector. Mastery of these operations leads to the correct simplification of the formula and ultimately, finding the correct distance.

Rationalizing the Denominator

Rationalizing the denominator is a technique used to eliminate radicals (such as square roots) from the bottom (denominator) of a fraction. The goal is to make the expression cleaner or more 'rational.' This process typically involves multiplying the numerator and the denominator of the fraction by the radical present in the denominator.

In our example, the distance formula gives us \(D = \frac{3}{\sqrt{22}}\), so we multiply the top and bottom by \(\sqrt{22}\) to get \(D = \frac{3\sqrt{22}}{22}\). This step does not change the value of the distance but makes the expression more standard and often easier to understand. It is an especially useful process when further mathematical manipulation is required or if a numerical approximation is sought.