Problem 40 - Thomas Calculus

Question

In Exercises \(39-44,\) find the distance from the point to the plane. $$ (0,0,0), \quad 3 x+2 y+6 z=6 $$

Step-by-Step Solution

Answer

The distance from the point to the plane is \(\frac{6}{7}\).

Step 1: Understand the Plane Equation

The equation of the plane is given as \(3x + 2y + 6z = 6\). From this equation, we can identify that the normal vector to the plane is \(\mathbf{n} = \langle 3, 2, 6 \rangle\).

Step 2: Understand the Point

The point given is \(P(0,0,0)\). We need to find the distance from this point \(P\) to the plane defined by the equation.

Step 3: Use the Distance Formula

The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is given by the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] Here, \(A = 3\), \(B = 2\), \(C = 6\), \(D = -6\), and the point is \((0,0,0)\).

Step 4: Substitute Values into the Formula

Substitute the values into the distance formula: \[ d = \frac{|3\cdot 0 + 2\cdot 0 + 6\cdot 0 + (-6)|}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{|-6|}{\sqrt{9 + 4 + 36}} \] This simplifies to: \[ d = \frac{6}{\sqrt{49}} \]

Step 5: Calculate the Distance

Calculate the square root and division: \( \sqrt{49} = 7 \), so \[ d = \frac{6}{7} \] Thus, the distance from the point to the plane is \(\frac{6}{7}\).

Key Concepts

Plane Equation

A plane in three-dimensional geometry is a flat, two-dimensional surface that stretches infinitely along its length and width. When we describe a plane mathematically, we use a plane equation, which is usually given in the form \(Ax + By + Cz = D\). This formula includes the variables \(x\), \(y\), and \(z\), which represent coordinates in 3D space, and the constants \(A\), \(B\), \(C\), and \(D\). These constants define the orientation and position of the plane. In our problem, the plane equation is \(3x + 2y + 6z = 6\). This tells us how the plane slices through the three-dimensional space. Understanding the plane equation is essential because it forms the foundation for calculating geometric relationships, like distances, within the 3D space. By comprehending this formula, you can see how a plane interacts with the rest of the space.

Normal Vector

The normal vector is a key concept in understanding a plane's orientation. It is a vector that is perpendicular to the plane. This means it points directly outward and is orthogonal to every direction lying in the plane. That vector helps in determining various properties of the plane, such as its angle of inclination. In our example, the normal vector is derived from the coefficients of \(x\), \(y\), and \(z\) in the plane equation. For the equation \(3x + 2y + 6z = 6\), the normal vector is \(\mathbf{n} = \langle 3, 2, 6 \rangle\). This vector not only characterizes the plane's direction but is also involved in computing distances to the plane. It's crucial to identify and use the normal vector when performing calculations involving the plane.

Distance Formula for Point to Plane

Finding the distance from a point to a plane in three-dimensional space involves using a specific formula. This formula calculates the shortest linear path between a given point and a plane, effectively giving us how far the point lies vertically from the plane. The distance formula is expressed as:

• \(d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\)

To apply this formula correctly, you need to substitute the individual coordinates of the point \((x_1, y_1, z_1)\) and the coefficients from the plane equation \(Ax + By + Cz + D = 0\). For instance, in our problem, the point is \((0,0,0)\), and using the plane equation \(3x + 2y + 6z = 6\), we substitute into the formula to find the distance. Through substitution and simplification, we find that the distance is \(\frac{6}{7}\). This formula is very useful in various applications where understanding spatial relationships is important.

Three-Dimensional Geometry

Three-dimensional geometry is a branch of mathematics that explores dimensions beyond the usual two axes—length and width—to include depth. It's all about understanding shapes, sizes, and the properties of three-dimensional objects. In 3D geometry, we work with coordinates \((x, y, z)\), which are points defined in space on three axes. This third dimension makes calculations and visualizations more complex but also more relevant to real-world scenarios.
By working within this framework, we can solve practical problems involving distances, angles, and intersections that occur in objects like cubes, spheres, and planes. In our problem, we used three-dimensional geometry to find the distance from a point to a plane. Mastering these concepts will help you to solve a vast array of problems in fields like engineering, physics, and computer graphics.