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Chapter 1: Problem 24

Find the distance between parallel planes \(5 x-2 y+z=6\) and \(5 x-2 y+z=-3\).

Short Answer

Expert verified
The distance between the planes is \(\frac{3 \sqrt{30}}{10}\).

Step by step solution

01

Identify Plane Equations

The two parallel planes are given by the equations:Plane 1: (Equation 1) \(5x - 2y + z = 6\)Plane 2: (Equation 2) \(5x - 2y + z = -3\).
02

Confirm Parallelism

Check if the normal vectors of the planes are the same, which confirms that the planes are parallel. The normal vector for both planes is \( \vec{n} = \langle 5, -2, 1 \rangle \), confirming that they are parallel.
03

Use the Distance Formula Between Parallel Planes

The formula to find the distance \(d\) between two parallel planes \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\) is given by:\[ d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} \].
04

Substitute Values Into Distance Formula

Substitute the given values:- \(D_1 = 6\) and \(D_2 = -3\),- The coefficients \(A = 5\), \(B = -2\), \(C = 1\) from the normal vector.The distance \(d\) can be calculated as:\[ d = \frac{|6 - (-3)|}{\sqrt{5^2 + (-2)^2 + 1^2}} \].
05

Simplify the Calculations

Calculate the numerator:\(|6 - (-3)| = |6 + 3| = 9\).Calculate the denominator:\(\sqrt{5^2 + (-2)^2 + 1^2} = \sqrt{25 + 4 + 1} = \sqrt{30}\).Thus, the distance \(d\):\[ d = \frac{9}{\sqrt{30}} \].
06

Final Calculation

Simplify \(\frac{9}{\sqrt{30}}\) by multiplying the numerator and denominator by \(\sqrt{30}\) to rationalize the denominator:\[ d = \frac{9 \times \sqrt{30}}{30} = \frac{3 \sqrt{30}}{10} \].

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Key Concepts

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

Normal Vectors
In geometry, the normal vector is a critical concept when dealing with planes. It essentially determines the direction of the plane in a three-dimensional space. For a plane defined by the equation \(Ax + By + Cz = D\), the normal vector is given by the coefficients of \(x\), \(y\), and \(z\), or \(\vec{n} = \langle A, B, C \rangle\).
For example, in our given plane equations \(5x - 2y + z = 6\) and \(5x - 2y + z = -3\), the normal vector \(\vec{n} = \langle 5, -2, 1 \rangle\) is derived from these coefficients.

Understanding normal vectors is essential because:
  • They provide the direction that a plane is "facing."
  • Two planes are parallel if and only if their normal vectors are parallel. This means the vectors are scalar multiples of each other.
  • If the normal vectors of two planes are identical, like in this problem, they are exactly parallel.
Parallel Planes
Parallel planes are planes that never intersect. They maintain a constant distance apart forever. This is similar to parallel lines in two dimensions but extended into three dimensions.
For instance, our provided planes \(5x - 2y + z = 6\) and \(5x - 2y + z = -3\) are parallel since their normal vectors are the same.

Key points to remember about parallel planes include:
  • Identical normal vectors assure the planes are parallel.
  • The concept is reliant on the orientations dictated by the normal vectors.
  • If their normal vectors are not scalar multiples of each other, the planes will eventually intersect.
  • Parallel planes can have different constant values (the \(D\) in \(Ax + By + Cz = D\)), indicating their distance from the origin.
Distance Formula
To find the distance between two parallel planes, we utilize a specific distance formula. For parallel planes defined by \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\), the formula is:\[d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}\]
This formula calculates the perpendicular distance between these planes.

In our example:
  • We have \(D_1 = 6\) and \(D_2 = -3\).
  • The coefficients from the normal vector \(\vec{n} = \langle 5, -2, 1 \rangle\): \(A = 5\), \(B = -2\), \(C = 1\).
Plug these into the formula:\[ d = \frac{|6 - (-3)|}{\sqrt{5^2 + (-2)^2 + 1^2}} = \frac{9}{\sqrt{30}}\]Furthermore, simplifying this by multiplying by \(\sqrt{30}\) results in:\[ d = \frac{3 \sqrt{30}}{10}\]This distance is how far apart, in a perpendicular line segment, these parallel planes are from each other.

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Most popular questions from this chapter

Consider \(\mathbf{r}(t)=\langle\sin t, \cos t, 2 t\rangle\) the position vector of a particle at time \(t \in[0,3]\), where the components of \(\mathbf{r}\) are expressed in centimeters and time is measured in seconds. Let \(\overrightarrow{O P}\) be the position vector of the particle after 1 sec a. Determine the velocity vector \(\mathbf{v}(1)\) of the particle after 1 sec. b. Find the scalar equation of the plane that is perpendicular to \(\mathbf{v}(1)\) and passes through point \(P\). This plane is called the normal plane to the path of the particle at point \(P\). c. Use a CAS to visualize the path of the particle along with the velocity vector and normal plane at point \(P\).

For the following exercises, the equations of two planes are given. Determine whether the planes are parallel, orthogonal, or neither. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer. Consider the plane of equation \(x-y-z-8=0\) a. Find the equation of the sphere with center \(C\) at the origin that is tangent to the given plane b. Find parametric equations of the line passing through the origin and the point of tangency.

Two forces, a vertical force of \(26 \mathrm{lb}\) and another of \(45 \mathrm{lb}\), act on the same object. The angle between these forces is \(55^{\circ}\). Find the magnitude and direction angle from the positive \(x\) -axis of the resultant force that acts on the object. (Round to two decimal places.)

A car is towed using a force of \(1600 \mathrm{~N}\). The rope used to pull the car makes an angle of \(25^{\circ}\) with the horizontal. Find the work done in towing the car \(2 \mathrm{~km}\). Express the answer in joules \((1 \mathrm{~J}=1 \mathrm{~N} \cdot \mathrm{m})\) rounded to the nearest integer.

For the following exercises, the equations of two planes are given. Determine whether the planes are parallel, orthogonal, or neither. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer. \(x-3 y+6 z=4,5 x+y-z=4\)
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