Question

In Exercises \(85-88,\) find the distance between the point and the plane. $$ \begin{array}{l} (0,0,0) \\ 2 x+3 y+z=12 \end{array} $$

Short answer

The short answer is, the distance from the point (0,0,0) to the plane 2x+3y+z=12 is \(\frac{12\sqrt{14}}{14}\)

Step-by-step solution

Identify Plane and Point

The plane is given by the equation 2x+3y+z=12, and the point is given as (0,0,0).

Apply Distance Formula

Apply the formula to calculate the distance from a point to a plane, which is \[ \frac{|Ax+By+Cz+D|}{\sqrt{A^2+B^2+C^2}} \]. Here, A=2, B=3, C=1, D=-12, and (x,y,z) is (0,0,0). Substituting these values, we get \[ \frac{|2*0 + 3*0 + 1*0-12|}{\sqrt{2^2+3^2+1^2}} \]

Calculate Distance

On calculation, the result within the absolute value becomes |-12| and the denominator becomes \(\sqrt{14}\), yielding \(\frac{12}{\sqrt{14}}\). This can be simplified by multiplying the numerator and denominator by \(\sqrt{14}\) to result in final answer of \(\frac{12\sqrt{14}}{14}\).

Key concepts

Coordinate Geometry

Coordinate geometry, often known as analytic geometry, is a field of mathematics that describes geometric shapes using a coordinate system and algegraic principles. This allows for a fusion of algebraic equations with geometric curves and figures.

In a three-dimensional space, any point can be represented by a triplet of numbers \((x, y, z)\), which correspond to its coordinates along the x-axis, y-axis, and z-axis, respectively. In the context of the given exercise, the plane is a flat, two-dimensional surface that extends infinitely in 3D space. It is defined by the equation \(2x + 3y + z = 12\).

The point in question is described as \((0,0,0)\), also known as the origin. It's crucial in coordinate geometry to consider the positions of both the point and the plane as they relate to one another in 3D space. This sets the stage for finding the shortest distance from the point to the plane, an essential application of coordinate geometry.

Distance Formula in 3D

The distance formula in 3D is a mathematical equation used to calculate the shortest distance between two objects in three-dimensional space. This formula is a fundamental concept not only in coordinate geometry but also in fields like physics and engineering.

When you're dealing with a point and a plane, the formula slightly changes. The shortest distance from a point to a plane is defined as being perpendicular to the plane itself. The equation for this distance is \[ \frac{|Ax + By + Cz + D|}{\(A^2+B^2+C^2\)} \], where \(A\), \(B\), and \(C\) are the coefficients that define the plane's orientation in 3D space and \(D\) is the plane's distance from the origin along the normal vector. Applying this to the provided exercise, you simply plug in the coefficients from the plane equation, and the coordinates of the point into this formula to calculate the distance.

Calculus

Calculus is often perceived as an advanced branch of mathematics, focused on the concepts of limits, functions, derivatives, and integrals, and it has a profound relationship with geometry. Although calculus might seem to overcomplicate the process of finding the distance from a point to a plane at first glance, it actually underpins many of the principles used.

For example, derivatives describe the slope of a function and can determine the rate of change in various applications. Understanding gradient, a vector consisting of partial derivatives, can be crucial when working with planes and distances in 3D space. In the context of our problem, calculus isn't necessarily employed to find the direct answer. However, the formula we use for the distance from a point to a plane arises from the principles of calculus, more specifically, from linearization and optimization problems.