Problem 15 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (1,5,-4), 3 x-y+2 z=6 $$

Step-by-Step Solution

Answer

The distance between the point and the plane is \(D=\frac{9\sqrt{14}}{14}\) units.

Step 1: Identify the coordinates of the point and coefficients/constant of the plane

The coordinates of the point, \((x_0, y_0, z_0)\), are given as (1, 5, -4). The coefficients a, b, c and constant d from the plane equation 3x - y + 2z = 6 are 3, -1, 2 and -6 respectively.

Step 2: Substitute the coordinates and constants into the distance formula

The distance D is given by the formula \[D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\]. Substituting our values from Step 1, we get \[D=\frac{\left|3*(1)+(-1)*(5)+2*(-4)+(-6)\right|}{\sqrt{3^{2}+(-1)^{2}+2^{2}}}\].

Step 3: Calculate the top and bottom parts of the fraction separately

On top, we have \(3*1 + (-1)*5 + 2*(-4) +(-6) = -9\), taking the absolute value, we get 9. On the bottom, we have \(\sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14}\).

Step 4: Divide the top by the bottom to get the distance

The formula, once simplified, results in \[D=\frac{9}{\sqrt{14}}\].

Step 5: Simplify the Result

To simplify this, we will rationalize the denominator by multiplying the numerator and the denominator each by \(\sqrt{14}\) giving the result in simplest radical form as \(D=\frac{9\sqrt{14}}{14}\).

Key Concepts

Distance formula in 3D geometry

Finding the distance between a point and a plane in three-dimensional geometry can be tackled using a specific formula. This formula originates from principles in vector mathematics and helps quantify how far separated two entities (a point and a plane) are in space.

• Consider a plane defined by the equation \(ax + by + cz + d = 0\).
• A point in 3D space is given by \((x_0, y_0, z_0)\).
• The distance \(D\) from this point to the plane is calculated using

\[D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}\]
The formula is derived by considering the plane's normal vector and how it projects perpendicularly to the point, allowing for an accurate linear measurement of distance. This approach ensures precision, as we're assessing the shortest path in the direction orthogonal to the plane, much like dropping a 'perpendicular' from the point to the surface.

Distance calculation

Using the distance formula requires careful substitution and arithmetic to ensure accuracy. Let's break down the calculation process using an example with actual numbers to understand each part of the formula:

• Suppose our point \((x_0, y_0, z_0)\) is (1, 5, -4), and the plane is defined by \(3x - y + 2z = 6\).
• First, identify the coefficients \(a=3\), \(b=-1\), \(c=2\), and constant \(d = -6\) (note that the plane equation is rewritten for standard form).
• Substitute these values into the distance formula:

\[D = \frac{|3(1) + (-1)(5) + 2(-4) + (-6)|}{\sqrt{3^2 + (-1)^2 + 2^2}}\]

• Calculate the numerator separately: \(3*1 - 1*5 + 2*(-4) - 6 = -9\), then take absolute value to get 9.
• Next, execute the denominator calculation: \(\sqrt{9 + 1 + 4} = \sqrt{14}\).
• Now, divide to find \(D\): \(\frac{9}{\sqrt{14}}\), which simplifies to \(\frac{9\sqrt{14}}{14}\) after rationalizing the denominator.

This clear sequence of steps ensures we follow the method methodically, minimizing calculation errors and improving understanding.

Three-dimensional coordinate geometry

Three-dimensional coordinate geometry is the study of geometric figures found in the 3D spatial environment. Understanding this is critical as it applies to real-world situations where objects do not exist merely in two planes but rather have depth.

• A point in 3D is represented by coordinates \((x, y, z)\), denoting its location with respect to three mutually perpendicular axes (usually labeled X, Y, and Z).
• Planes in 3D are continuous surfaces and can be expressed in the form \(ax + by + cz + d = 0\). This represents the set of points forming the plane.
• The distance between a point and plane provides insights into the position of the point relative to the plane, crucial for fields such as architecture, aviation, and computer graphics where spatial relationships must be fully understood.

By combining this understanding with algebraic expressions, three-dimensional coordinate geometry allows for powerful problem-solving tools, enhancing our ability to manipulate and analyze spatial figures.