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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$. $$ (-2,1,0), 2 x+5 y-z=20 $$

Short answer

The distance between the point (-2,1,0) and the plane \(2x+5y-z = 20\) is \( \frac{19}{\sqrt{30}} \)

Step-by-step solution

Identify the variables

From the given exercise, identify the coordinates of the point which are \(x_{0} = -2\), \(y_{0} = 1\), and \(z_{0} = 0\). Also, identify the coefficients of the equation representing the plane which are \(a = 2\), \(b = 5\), \(c = -1\), and \(d = -20\) (Note that the given equation is equal to zero)

Substitute the values into the formula

Next, substitute these known values into the formula to get: \[ D = \frac{\left| 2(-2) + 5(1) + -1(0) + -20 \right|} {\sqrt{2^2 + 5^2 + (-1)^2}} \]

Simplification

Now, simplify the equation. In the numerator, the expression becomes -4 + 5 + 0 - 20 which equals -19. In the denominator, the expression is \(\sqrt{4 + 25 + 1}\) which equals \(\sqrt{30}\). Therefore, the calculation becomes \[ D= \frac{|-19|}{\sqrt{30}} \]

Final Calculation

Finally, take the absolute value of the numerator and simplify if possible. The absolute value of -19 is 19, so the final answer for \(D\) becomes \[ D=\frac{19}{\sqrt{30}} \]

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that involves the study of geometric figures using a system of coordinates. This system, often represented as a grid with an x, y, and z-axis in three-dimensional space, allows us to locate the position of points and describe geometric shapes algebraically.

For example, the point \(x_0, y_0, z_0\) represents a specific location in space determined by its coordinates. Such a point can be related to various geometric entities, such as lines and planes, through equations that incorporate its coordinates. Understanding how to navigate this coordinate system and interpret the relationships between different elements is crucial in solving a wide array of problems in geometry and beyond.

Applying coordinate geometry to our original exercise, one must be able to identify the point \(x_0, y_0, z_0\) and comprehend its connection to the equation of a plane in a three-dimensional space to compute the distance between them. As we delve further into algebraic equations and plane equations, the interplay between these concepts becomes increasingly evident.

Algebraic Equations

Algebraic equations are mathematical statements that utilize algebra to express one or more unknown variables in terms of other numbers and variables. They can range from very simple, like \(x + 2 = 5\), to highly complex, involving multiple variables and powers.

When we solve an algebraic equation, we are finding the values of the variables that make the equation true. This involves manipulating the equation, often using basic operations of addition, subtraction, multiplication, and division, along with more advanced concepts such as factoring and employing the quadratic formula, where appropriate.

In the context of the distance between a point and a plane, we use algebraic equations to express the given elements and then solve for the desired quantity. By organizing the known values from the point's coordinates and the coefficients of the plane's equation, and then systematically substituting and simplifying, as shown in the step by step solution, we use algebra to achieve the final answer.

Plane Equation

A plane in three-dimensional space can be defined algebraically by an equation of the form \(ax + by + cz + d = 0\), where \(a\), \(b\), and \(c\) are coefficients that represent the orthogonal (perpendicular) direction vector to the plane, and \(d\) is the constant term. This equation is known as the general form of the equation of a plane.

Any point \(x, y, z\) that satisfies this equation lies on the plane, and any point that does not satisfy it lies elsewhere in space. The beauty of this expression is that it provides a succinct way to describe an infinitely large, two-dimensional flat surface within the three-dimensional coordinate space.

Understanding the plane equation is central to determining the distance from a point to the plane. This concept makes it possible to create a formula that takes the point's coordinates and the plane's normal vector to calculate this shortest distance. As seen in the exercise, once the values for \(a, b, c,\) and \(d\) are known, along with the coordinates of the point, it becomes an algebraic challenge to resolve the actual distance using the derived formula.