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Chapter 11: Problem 75

Find the distance between the parallel planes \(-3 x+2 y+z=9\) and \(6 x-4 y-2 z=19\).

Short Answer

Expert verified
The distance between the planes is \(\frac{37\sqrt{14}}{28}\).

Step by step solution

01

Check for Parallelism

Ensure the given planes are parallel. The equations of the planes are given by:Plane 1: \(-3x + 2y + z = 9\)Plane 2: \(6x - 4y - 2z = 19\)The normal vector to Plane 1 is \((-3, 2, 1)\) and to Plane 2 is \((6, -4, -2)\). Since \(6, -4, -2 = -2(-3, 2, 1)\), the planes are indeed parallel.
02

Identify Form

Convert the second plane equation to have the same normal vector sign as the first by dividing by \(-2)\): Divide: \(-\frac{1}{2}(6x - 4y - 2z = 19)\), resulting in: \(3x - 2y - z = -\frac{19}{2}\), matching the direction of Plane 1's normal vector.
03

Compute Distance Formula Parameters

Use the formula for the distance \(d\) between two parallel planes \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\):\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]Here, \(D_1 = 9\) and \(D_2 = -\frac{19}{2}\), using \(A = -3\), \(B = 2\), and \(C = 1\).
04

Calculate the Distance

Substitute the values into the distance formula:\[ d = \frac{|9 - (-\frac{19}{2})|}{\sqrt{(-3)^2 + 2^2 + 1^2}} = \frac{|9 + \frac{19}{2}|}{\sqrt{14}} \]Simplify the numerator: \(9 + \frac{19}{2} = \frac{18}{2} + \frac{19}{2} = \frac{37}{2}\).Calculate the distance: \[ d = \frac{\frac{37}{2}}{\sqrt{14}} = \frac{37}{2\sqrt{14}} \]Multiply numerator and denominator by \(-14\)\ to eradicate the square root: \[ d = \frac{37\sqrt{14}}{28} \].

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Key Concepts

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

Normal vector
In three-dimensional geometry, the normal vector of a plane is a vector that is perpendicular to the surface of that plane. This concept is crucial in understanding the orientation and alignment of geometric planes. By examining the coefficients of the plane equation, you can derive this normal vector. For a plane given by the equation \(Ax + By + Cz = D\), the normal vector is \((A, B, C)\).

The normal vector tells us about the orientation of the plane. For instance, in the exercise's first plane equation \(-3x + 2y + z = 9\), the normal vector is \((-3, 2, 1)\). Similarly, for the second plane \(6x - 4y - 2z = 19\), the initial normal vector can be scaled or adjusted as needed for calculations. These vectors help determine if two planes are parallel, orthogonal, or intersecting, based on their geometric properties.

Moreover, normalization often involves scaling a vector to simplify comparisons or computations. By comparing normal vectors, we can establish geometric relations among surfaces, laying the groundwork for more complex operations such as calculating distances or angles.
Distance formula
The distance formula for parallel planes is a specific equation that allows you to calculate the shortest distance between two planes that never intersect. Given two parallel planes with shared normal vector coefficients \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\), the distance \(d\) is computed using the formula:

\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]

This formula calculates the Euclidean distance, which is the shortest path, from a point on one plane perpendicular to the other plane along their mutually shared normal vector. The numerator expresses the absolute difference of the plane constants \(D_1\) and \(D_2\). The denominator represents the magnitude of the normal vector \((A, B, C)\), ensuring the correct scalar distance computation between these layers in space.

Application of this formula requires ensuring the planes share identical normal vector orientations. This scenario often arises in mathematical problems involving symmetry, construction projects, and various engineering applications where precise spatial relationships are critical.
Parallelism in geometry
Parallelism in geometry refers to the condition when two or more lines or planes run side by side equidistant from each other and never meet, no matter how far they are extended. For 3D planes, parallelism happens when they have identical or scalar multiples of their normal vectors, indicating that they share the same orientation in space.

Considerations of parallelism extend to analyzing characteristics such as distance measurements, collisional checks in virtual simulations, and structural alignments in architecture. In the exercise given, by confirming the normal vectors of the planes are scalar multiples, we verify the parallelism, a foundational step required before computing any distances.

When diving deeper, parallelism is not only a static property. It translates into dynamic systems, influencing overall geometry in physics and engineering, such as ensuring elements in designs remain consistently aligned for functionality or visual balance.

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Most popular questions from this chapter

EXPL 48. In this exercise you will derive Kepler's First Law, that planets travel in elliptical orbits. We begin with the notation. Place the coordinate system so that the sun is at the origin and the planet's closest approach to the sun (the perihelion) is on the positive \(x\) -axis and occurs at time \(t=0\). Let \(\mathbf{r}(t)\) denote the position vector and let \(r(t)=\|\mathbf{r}(t)\|\) denote the distance from the sun at time \(t\). Also, let \(\theta(t)\) denote the angle that the vector \(\mathbf{r}(t)\) makes with the positive \(x\) -axis at time \(t\). Thus, \((r(t), \theta(t))\) is the polar coordinate representation of the planet's position. Let \(\mathbf{u}_{1}=\mathbf{r} / r=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j} \quad\) and \(\quad \mathbf{u}_{2}=(-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}\) Vectors \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal unit vectors pointing in the directions of increasing \(r\) and increasing \(\theta\), respectively. Figure 12 summarizes this notation. We will often omit the argument \(t\), but keep in mind that \(\mathbf{r}, \theta, \mathbf{u}_{1}\), and \(\mathbf{u}_{2}\) are all functions of \(t .\) A prime in. dicates differentiation with respect to time \(t\). (a) Show that \(\mathbf{u}_{1}^{\prime}=\theta^{\prime} \mathbf{u}_{2}\) and \(\mathbf{u}_{2}^{\prime}=-\theta^{\prime} \mathbf{u}_{1}\). (b) Show that the velocity and acceleration vectors satisfy $$ \begin{array}{l} \mathbf{v}=r^{\prime} \mathbf{u}_{1}+r \theta^{\prime} \mathbf{u}_{2} \\ \mathbf{a}=\left(r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2}\right) \mathbf{u}_{1}+\left(2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime}\right) \mathbf{u}_{2} \end{array} $$ (c) Use the fact that the only force acting on the planet is the gravity of the sun to express a as a multiple of \(\mathbf{u}_{1}\), then explain how we can conclude that $$ \begin{aligned} r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2} &=\frac{-G M}{r^{2}} \\ 2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime} &=0 \end{aligned} $$ (d) Consider \(\mathbf{r} \times \mathbf{r}^{\prime}\), which we showed in Example 8 was a constant vector, say D. Use the result from (b) to show that \(\mathbf{D}=r^{2} \theta^{\prime} \mathbf{k} .\) (e) Substitute \(t=0\) to get \(\mathbf{D}=r_{0} v_{0} \mathbf{k}\), where \(r_{0}=r(0)\) and \(v_{0}=\|\mathbf{v}(0)\|\). Then argue that \(r^{2} \theta^{\prime}=r_{0} v_{0}\) for all \(t\). (f) Make the substitution \(q=r^{\prime}\) and use the result from (e) to obtain the first-order (nonlinear) differential equation in \(q\) : $$ q \frac{d q}{d r}=\frac{r_{0}^{2} v_{0}^{2}}{r^{3}}-\frac{G M}{r^{2}} $$ (g) Integrate with respect to \(r\) on both sides of the above equation and use an initial condition to obtain $$ q^{2}=2 G M\left(\frac{1}{r}-\frac{1}{r_{0}}\right)+v_{0}^{2}\left(1-\frac{r_{0}^{2}}{r^{2}}\right) $$ (h) Substitute \(p=1 / r\) into the above equation to obtain $$ \frac{r_{0}^{2} v_{0}^{2}}{\left(\theta^{\prime}\right)^{2}}\left(\frac{d p}{d t}\right)^{2}=2 G M\left(p-p_{0}\right)+v_{0}^{2}\left(1-\frac{p^{2}}{p_{0}^{2}}\right) $$

A dog is running counterclockwise around the circle \(x^{2}+y^{2}=400\) (distances in feet). At the point \((-12,16)\), it is running at 10 feet per second and is speeding up at 5 feet per second per second. Express its acceleration \(\mathbf{a}\) at the point first in terms of \(\mathbf{T}\) and \(\mathbf{N}\), and then in terms of \(\mathbf{i}\) and \(\mathbf{j}\)

Find the equation of the plane each of whose points is equidistant from \((-2,1,4)\) and \((6,1,-2)\).

find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\) $$ \mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j} ; t_{1}=\pi / 6 $$

Find the equation of the surface that results when the curve \(4 x^{2}-3 y^{2}=12\) in the \(x y\) -plane is revolved about the \(x\) -axis.
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