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Chapter 11: Problem 40

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

Short Answer

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

Step by step solution

01

Identify the Point and the Plane Equation

We have a point given as \((0,0,0)\) and a plane described by the equation \(3x + 2y + 6z = 6\). Our task is to find the distance from the point to this plane.
02

Write the Distance Formula

The distance from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is given by the formula: \[ \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] where \(A = 3\), \(B = 2\), \(C = 6\), \(D = -6\).
03

Substitute Values into the Formula

Substitute the values and point into the distance formula: \(x_1 = 0\), \(y_1 = 0\), \(z_1 = 0\). The formula becomes: \[ \frac{|3(0) + 2(0) + 6(0) - 6|}{\sqrt{3^2 + 2^2 + 6^2}} \]
04

Simplify the Numerator

Calculate the numerator: \[ |3(0) + 2(0) + 6(0) - 6| = |-6| = 6 \]
05

Simplify the Denominator

Calculate the denominator: \[ \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]
06

Calculate the Distance

Apply the values to the formula: \[ \frac{6}{7} \] Thus, the distance from the point \((0,0,0)\) to the plane \(3x + 2y + 6z = 6\) is \(\frac{6}{7}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plane Equation
A plane in 3D geometry is defined by an equation that contains variables and coefficients. In this exercise, the plane equation given is:
  • The equation: \( 3x + 2y + 6z = 6 \).
  • The role of coefficients \( A = 3 \), \( B = 2 \), and \( C = 6 \) is vital. These determine the orientation of the plane in space.
A plane's equation can also be expressed in standard form: \( Ax + By + Cz + D = 0 \). By rearranging the terms of our plane equation, we get:
  • \( 3x + 2y + 6z - 6 = 0 \), where \( D = -6 \).
This rearrangement helps to apply formulas, like the distance formula, correctly. Planes in 3D allow us to analyze spatial relationships, crucial for understanding distances and interactions with points in space.
Distance Formula
To calculate the shortest distance from a point to a plane, we utilize a specific distance formula. This is crucial for understanding how points and planes in 3D interact. Let's take a closer look:
  • The formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] is used to find the perpendicular distance from a point \((x_1, y_1, z_1)\) to the plane \( Ax + By + Cz + D = 0 \).
  • In our exercise, the point is \((0,0,0)\), and values from the plane equation \((A = 3, B = 2, C = 6, D = -6)\) are substituted.
Inserting these values simplifies the formula, simplifying the process of finding the distance. The formula's use of absolute values in the numerator ensures positive distance values, reflecting the real-world spatial separation.Applying this knowledge, students can solve a broad range of geometric problems involving distances to planes.
3D Geometry
3D geometry is an extension of what we learn in basic geometry, adding depth to known concepts. Here, we work not just on a plane but within three-dimensional space, which includes:
  • Horizontal and vertical planes, as well as the third dimension of depth.
  • Understanding 3D requires visualizing equations and points in space, like how a point \((0,0,0)\) relates to a plane \(3x+2y+6z=6\).
In the context of our problem, 3D geometry allows us to seek the shortest route 鈥� the perpendicular distance 鈥� from a point to a plane. Visualizing these spaces helps in comprehending how objects are placed relative to one another and the respective distances involved.The mental model of 3D space is crucial for various fields, like physics, engineering, and computer graphics, all of which rely on principles of 3D geometry to solve real-world problems efficiently.

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Most popular questions from this chapter

Use a CAS to plot the surfaces in Exercises. Identify the type of quadric surface from your graph. $$\frac{y^{2}}{16}=1-\frac{x^{2}}{9}+z$$

Which of the following are always true, and which are not always true? Give reasons for your answers. a. \(\mathbf{u} \cdot \mathbf{v}=\mathbf{v} \cdot \mathbf{u}\) b. \(\mathbf{u} \times \mathbf{v}=-(\mathbf{v} \times \mathbf{u})\) c. \((-\mathbf{u}) \times \mathbf{v}=-(\mathbf{u} \times \mathbf{v})\) d. \((c \mathbf{u}) \cdot \mathbf{v}=\mathbf{u} \cdot(c \mathbf{v})=c(\mathbf{u} \cdot \mathbf{v}) \quad \text { (any number } c)\) e. \(c(\mathbf{u} \times \mathbf{v})=(c \mathbf{u}) \times \mathbf{v}=\mathbf{u} \times(c \mathbf{v}) \quad \text { (any number } c)\) f. \(\mathbf{u} \cdot \mathbf{u}=|\mathbf{u}|^{2}\) g. \((\mathbf{u} \times \mathbf{u}) \cdot \mathbf{u}=0\) h. \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u}=\mathbf{v} \cdot(\mathbf{u} \times \mathbf{v})\)

Find the volume of a parallelepiped with one of its eight vertices at \(A(0,0,0)\) and three adjacent vertices at \(B(1,2,0), C(0,-3,2)\) and \(D(3,-4,5)\).

How can you tell when two planes \(A_{1} x+B_{1} y+C_{1} z=D_{1}\) and \(A_{2} x+B_{2} y+C_{2} z=D_{2}\) are parallel? Perpendicular? Give reasons for your answer.

Suppose \(L_{1}\) and \(L_{2}\) are disjoint (nonintersecting) nonparallel lines. Is it possible for a nonzero vector to be perpendicular to both \(L_{1}\) and \(L_{2} ?\) Give reasons for your answer.
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