Problem 17 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (1,0,-1), 2 x-4 y+3 z=12 $$

Step-by-Step Solution

Answer

The distance D between the point (1,0,-1) and the plane \(2x - 4y + 3z = 12\) is \(\frac{9}{\sqrt{29}}\).

Step 1: Determine the coefficients of the plane equation and the coordinates of the point

The equation for the plane is \(2x - 4y + 3z = 12\), from which you can identify the coefficients as \(a = 2\), \(b = -4\), and \(c = 3\). The constant \(d\) can be identified by bringing 12 over to the left side of the equation, giving \(d = -12\). The given point's coordinates are \((1, 0, -1)\).

Step 2: Substitute the values into the distance formula

Substitute \((x_0 = 1, y_0 = 0, z_0 = -1)\) and \((a = 2, b = -4, c = 3, d = -12)\) into the formula for \(D\). Thus, \(D = \frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\), simplifies to \(D = \frac{\left|2*1+(-4)*0+3*(-1) - 12\right|}{\sqrt{2^{2}+(-4)^{2}+3^{2}}} = \frac{\left|-9\right|}{\sqrt{29}}\).

Step 3: Simplify the equation

Presenting the final simplified equation, \(D = \frac{9}{\sqrt{29}}\).

Key Concepts

Plane Equation Coefficients

The equation of a plane in three-dimensional space is generally expressed as \( ax + by + cz + d = 0 \), where \( a \), \( b \), and \( c \) are the coefficients that determine the orientation of the plane in space. These coefficients are crucial as they represent the components of a normal vector to the plane. Understanding this helps in realizing how the plane is tilted or oriented along the x, y, and z axes.
For instance, in the given plane equation \( 2x - 4y + 3z = 12 \), the coefficients are \( a = 2 \), \( b = -4 \), and \( c = 3 \). These coefficients can directly be associated with the direction ratios of a normal vector to the plane, \( \mathbf{n} = (2, -4, 3) \). By shifting the constant \( 12 \) across the equals sign, the equation transforms to \( 2x - 4y + 3z - 12 = 0 \), allowing us to identify \( d \) as \( -12 \). The value \( d \) affects the position of the plane along the normal direction but does not influence its orientation.

Distance Formula

The distance formula is a useful tool in coordinate geometry, especially for finding how far a point is from a plane. The formula is expressed as:
\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \] This equation calculates the perpendicular distance from a point \((x_0, y_0, z_0)\) to the plane \( ax + by + cz + d = 0 \). It is crucial to follow these steps to apply the formula:

• Identify and plug in the coefficients \( a, b, c, \text{and } d \) from the plane equation.
• Insert the coordinates \( (x_0, y_0, z_0) \) of the point in question.
• Calculate the numerator by substituting these values into the expression \( |ax_0 + by_0 + cz_0 + d| \).
• For the denominator, compute \( \sqrt{a^2 + b^2 + c^2} \), reflecting the length of the normal vector to the plane.

It simplifies comparison of distances across different points and planes, thus enhancing our grasp over spatial geometrical relations.

Coordinate Geometry

Coordinate geometry, also known as analytical geometry, allows us to interpret geometric problems using algebraic equations. It serves as the bridge between algebra and geometry, enabling the translation of visual problems into solvable equations.
This involves defining positions of points, lines, and planes in a coordinate system. For example, to locate a point (such as \((1, 0, -1)\)) within space, you use the point's coordinates. Similarly, planes, like the one given by \( 2x - 4y + 3z = 12 \), are represented by equations derived from the plane's normal vector and any point on the plane.
Coordinate geometry enhances our understanding by providing formulas, such as the distance formula, that quantify spatial relationships.

• Helps describe the position and movement within a 3D space.
• Facilitates problem-solving for distances, angles, and intersections in geometric structures.
  
• Makes it easier to plot complex shapes and analyze spatial relationships between them.

By mastering coordinate geometry, students gain essential skills for resolving both theoretical and practical geometry challenges.