Problem 16 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (3,2,1), x-y+2 z=4 $$

Step-by-Step Solution

Answer

The distance \(D\) between point \((3,2,1)\) and the plane \(x - y + 2z = 4\) is \(\frac{1}{\sqrt{6}}\)

Step 1: Identification of Variables

First, identify the variables from the exercise. The plane equation given is \(x - y + 2z = 4\). From this, we can infer that \(a = 1\), \(b = -1\), \(c = 2\) and \(d = -4\). The point given is \((3,2,1)\) from where \(x_0 = 3\), \(y_0 = 2\) and \(z_0 = 1\).

Step 2: Application of the Formula

Now, plug these values into the distance formula \[D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\] The numerator of this formula will be calculated as \(1 \cdot 3 -1 \cdot 2 + 2 \cdot 1 -4 = -1\). However, since the absolute value bars mean we consider only the magnitude, or the positive value, we drop the negative sign, making it 1. The denominator will be calculated as \(\sqrt{1^{2} + (-1)^{2} + 2^{2}} = \sqrt{6}\).

Step 3: Calculation of the Distance

Now calculate the ratio which will give the distance \(D\). The distance between the point and plane \[D = \frac{1}{\sqrt{6}}\]

Key Concepts

Plane Equation

The equation of a plane in three-dimensional space is one of the fundamental concepts in vector algebra and analytical geometry. It is generally represented as a linear equation in the form:
\[ax + by + cz + d = 0\]
where \(a\), \(b\), and \(c\) are coefficients that create a vector perpendicular to the plane known as the normal vector, and \(d\) is the scalar constant. The normal vector is key in defining the orientation of the plane. Every point \((x, y, z)\) that satisfies the equation lies on the plane. For the exercise, the plane equation is \(x - y + 2z = 4\), which can be rewritten to the standard form by subtracting 4 from both sides, giving us \(x - y + 2z - 4 = 0\).
When solving problems involving planes, it's crucial to correctly identify these coefficients. For example, from the plane equation given in the exercise, the coefficients \(a = 1\), \(b = -1\), and \(c = 2\) are easily identified, and \(d\) is -4 after rearranging the equation to match the standard form. Understanding these components of the plane equation allows us to tackle a variety of geometrical problems, such as finding the distance from a point to the plane.

Distance Formula

The distance formula comes into play when calculating the shortest distance between a point and a plane in 3D space. This shortest distance is always along the line that is perpendicular to the plane, intersecting with the given point and the plane. The formula for finding this distance \(D\) from a point with coordinates \((x_{0}, y_{0}, z_{0})\) to a plane defined by the plane equation \(ax + by + cz + d = 0\) is given by:
\[D = \frac{|ax_{0} + by_{0} + cz_{0} + d|}{\sqrt{a^2 + b^2 + c^2}}\]
This formula can be derived using vector algebra and the concept of the dot product. The numerator represents the absolute value of the signed distance from the point to the plane, ensuring we get a non-negative result. The denominator normalizes this value by the length of the normal vector to the plane. It is essential when applying the distance formula to use the absolute value to account for the direction of the displacement. This makes sure that the distance is always a positive value, as required in the context of geometric measurements.
For the given problem, we substitute the coordinates of the point \((3, 2, 1)\) and the coefficients of the plane equation into the formula to calculate the distance. Following the steps accurately guarantees us the shortest distance between the point and the plane in question.

Vector Algebra

Vector algebra is the branch of mathematics that deals with operations performed on vectors. Vectors are entities that have both magnitude and direction, pivotal in representing quantities in physics and engineering. In the context of the distance between a point and a plane, vector algebra helps us understand the geometrical concepts involved.
Crucial vector operations include addition, subtraction, scalar multiplication, and notably, the dot product, which is used to determine the angle between vectors or project one vector onto another. In solving the distance problem, the dot product tells us how far the point lies from the origin along the direction of the normal vector to the plane.
Vector algebra simplifies the complex geometrical problems by translating them into algebraic calculations. To better grasp vector algebra concepts, it's helpful to visualize vectors as directed line segments and operations as combining these segments following the rules of vector algebra. The vector perpendicular to the plane in the exercise, the normal vector, for instance, is the key to unlocking the distance formula. Without understanding vectors and their properties, it would be much harder to interpret this and other spatial relationships in three-dimensional geometry.