Problem 29
Question
Find the given distances. The distance between the parallel planes \(x+y+z=0\) and $$ (x-2)+(y-3)+(z+4)=0 $$
Step-by-Step Solution
Verified Answer
The distance between the planes is \(3\sqrt{3}\).
1Step 1: Identify the Equations of the Planes
The equations of the parallel planes are given as \(x+y+z=0\) and \(x-2+y-3+z+4=0\). Simplify the second equation: it becomes \(x+y+z-9=0\). So, the equations can be written as \(x+y+z=0\) and \(x+y+z=9\).
2Step 2: Verify the Normal Vector
Both planes have the form \(x+y+z = d\), which indicates that the normal vector to both planes is \(\mathbf{n} = [1, 1, 1]\). This means the planes are indeed parallel since they share the same normal vector.
3Step 3: Formula for Distance 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:\[D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\]Identify \(d_1 = 0\) and \(d_2 = 9\), and \(a = b = c = 1\).
4Step 4: Substitute Values into the Formula
Insert the known values into the distance formula:\[D = \frac{|9 - 0|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{9}{\sqrt{3}}\]
5Step 5: Simplify the Expression
Simplify the expression:\[D = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}\]
6Step 6: Final Answer
The distance between the parallel planes \(x+y+z=0\) and \(x+y+z=9\) is \(3\sqrt{3}\).
Key Concepts
Parallel PlanesNormal VectorDistance Formula for PlanesSimplifying Equations
Parallel Planes
Parallel planes are fascinating as they extend infinitely without ever intersecting. Imagine two perfectly flat sheets of paper positioned alongside one another, always equidistant and never meeting or diverging. In mathematics, parallel planes are represented by linear equations that share identical normal vectors. This means the coefficients in front of the variables are the same.
For example, looking at the equations of our given planes: \(x+y+z=0\) and \(x+y+z=9\). Both share the same structure, ensuring they are parallel. They have equal normal vectors despite having different constant terms. This difference in constant values simply shifts the planes apart along a direction perpendicular to their surface, but doesn’t alter their infinite span.
For example, looking at the equations of our given planes: \(x+y+z=0\) and \(x+y+z=9\). Both share the same structure, ensuring they are parallel. They have equal normal vectors despite having different constant terms. This difference in constant values simply shifts the planes apart along a direction perpendicular to their surface, but doesn’t alter their infinite span.
Normal Vector
A normal vector is crucial in understanding the orientation of a plane. It is a vector that points perpendicular to the surface of a plane. Essentially, it helps determine the plane's tilt in three-dimensional space.
For any plane expressed as \(ax+by+cz=d\), the normal vector is given by \[\mathbf{n} = [a, b, c]\]. In our exercise, the normal vector for both planes is \[\mathbf{n} = [1, 1, 1]\], meaning these vectors point in the same direction. This is what confirms that the planes are parallel. Understanding normal vectors can help visualize how planes relate to each other in space, whether they are parallel or at an angle.
For any plane expressed as \(ax+by+cz=d\), the normal vector is given by \[\mathbf{n} = [a, b, c]\]. In our exercise, the normal vector for both planes is \[\mathbf{n} = [1, 1, 1]\], meaning these vectors point in the same direction. This is what confirms that the planes are parallel. Understanding normal vectors can help visualize how planes relate to each other in space, whether they are parallel or at an angle.
Distance Formula for Planes
Calculating the distance between two parallel planes is straightforward once you know they have the same normal vector. The distance formula between two parallel planes \(ax+by+cz=d_1\) and \(ax+by+cz=d_2\) is:
Using the distance formula provides a clear method to measure separation with ease, even in the complex space of three dimensions.
- \[D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\]
Using the distance formula provides a clear method to measure separation with ease, even in the complex space of three dimensions.
Simplifying Equations
Simplifying equations is a vital skill in algebra and geometry. It often makes challenging problems more approachable by reducing them to their core essence. In our problem, the second plane's equation started as \(x-2+y-3+z+4=0\).
Through simplification, we combined like terms to reformulate it as \(x+y+z=9\). By handling the arithmetic operations, we streamlined the equation, aligning it with the \(x+y+z=0\) plane's format. Consequently, the adjustments offer clarity and enable effortless parallel plane comparisons.
Simplifying is not just about making equations tidy; it's about enabling patterns to emerge, solving equations more efficiently, and understanding relationships among various mathematical elements.
Through simplification, we combined like terms to reformulate it as \(x+y+z=9\). By handling the arithmetic operations, we streamlined the equation, aligning it with the \(x+y+z=0\) plane's format. Consequently, the adjustments offer clarity and enable effortless parallel plane comparisons.
Simplifying is not just about making equations tidy; it's about enabling patterns to emerge, solving equations more efficiently, and understanding relationships among various mathematical elements.
Other exercises in this chapter
Problem 28
Verify, from Key Idea 10.2.1, that $$ \vec{u}=\langle\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta\rangle $$ is a unit vector for all angles \
View solution Problem 28
Sketch the quadric surface. \(z^{2}=x^{2}+\frac{y^{2}}{4}\)
View solution Problem 29
Find the area of the triangle with the given vertices. Vertices: (1,1),(1,3) and (2,2) .
View solution Problem 29
Vectors \(\vec{u}\) and \(\vec{v}\) are given. Write \(\vec{u}\) as the sum of two vectors, one of which is parallel to \(\vec{v}\) and one of which is perpendi
View solution