Chapter 11: Problem 16
Find the equation of the plane through \((0,0,2)\) that is parallel to the plane \(x+y+z=1\).
Short Answer
Expert verified
The equation of the plane is \(x + y + z = 2\).
Step by step solution
01
Understand the problem
We need to find the equation of a plane that passes through the point \((0,0,2)\) and is parallel to a given plane with the equation \(x+y+z=1\).
02
Identify the normal vector
The normal vector to a plane in the form \(Ax + By + Cz = D\) is \((A, B, C)\). For the plane \(x+y+z = 1\), the normal vector is \((1, 1, 1)\).
03
Write the equation of the parallel plane
A plane parallel to another will have the same normal vector. So the equation of our desired plane will have the form \(x + y + z = k\), where \(k\) is a constant.
04
Substitute the given point into the plane equation
Substitute the point \((0, 0, 2)\) into the plane equation \(x + y + z = k\). This gives \(0 + 0 + 2 = k\), so \(k=2\).
05
State the final equation
With \(k=2\), the equation of the desired plane is \(x + y + z = 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Vector
In the realm of 3D geometry, the normal vector is a crucial element when dealing with planes. Simply put, a normal vector is a vector that is perpendicular to a plane. It is what helps define the orientation of the plane in three-dimensional space. When you have a plane equation in the format of \(Ax + By + Cz = D\), the coefficients \((A, B, C)\) represent the normal vector. These numbers tell you the direction along which the surface would tilt in 3D space.
Knowing the normal vector is incredibly useful. It allows you to determine how planes are oriented relative to each other. Since the normal vector is perpendicular to the plane, it acts as a unique identifier for the plane's tilts and angles. For the plane given by the equation \(x + y + z = 1\), the normal vector is \((1, 1, 1)\).
This normal vector indicates that for every unit the plane moves in the x-direction, it moves equally in the y and z-directions. This understanding makes it easier to tackle problems involving parallel planes, intersections, and angles. So, always start by identifying the normal vector when working with plane equations.
Knowing the normal vector is incredibly useful. It allows you to determine how planes are oriented relative to each other. Since the normal vector is perpendicular to the plane, it acts as a unique identifier for the plane's tilts and angles. For the plane given by the equation \(x + y + z = 1\), the normal vector is \((1, 1, 1)\).
This normal vector indicates that for every unit the plane moves in the x-direction, it moves equally in the y and z-directions. This understanding makes it easier to tackle problems involving parallel planes, intersections, and angles. So, always start by identifying the normal vector when working with plane equations.
Parallel Planes
Planes that are parallel in three-dimensional space share a remarkable characteristic: they have the same normal vector. This implies that parallel planes never intersect with one another, no matter how far they extend in space. In simple terms, parallel planes maintain a consistent distance between each other.
When you need to find a plane that is parallel to a given plane, focus on matching their normal vectors. Since the normal vector for \(x + y + z = 1\) is \((1, 1, 1)\), any plane parallel to it will also have a normal vector of \((1, 1, 1)\).
The general form of the equation of any plane parallel to the given one will be \(x + y + z = k\), where \(k\) is a constant. This constant determines a specific plane parallel to the original but at a different position in space.
In our problem, by incorporating the point \((0,0,2)\) into the equation, we find that \(k=2\). Therefore, the plane equation \(x + y + z = 2\) is parallel to the original plane \(x + y + z = 1\). This confirms the principle that parallel planes share a normal vector but differ in their distance from the origin.
When you need to find a plane that is parallel to a given plane, focus on matching their normal vectors. Since the normal vector for \(x + y + z = 1\) is \((1, 1, 1)\), any plane parallel to it will also have a normal vector of \((1, 1, 1)\).
The general form of the equation of any plane parallel to the given one will be \(x + y + z = k\), where \(k\) is a constant. This constant determines a specific plane parallel to the original but at a different position in space.
In our problem, by incorporating the point \((0,0,2)\) into the equation, we find that \(k=2\). Therefore, the plane equation \(x + y + z = 2\) is parallel to the original plane \(x + y + z = 1\). This confirms the principle that parallel planes share a normal vector but differ in their distance from the origin.
Point on Plane
Having a point involved in the equation of a plane offers a direct path to identifying the exact plane in the infinite set of planes parallel to a given one. Once you've determined the form \(x + y + z = k\) for a set of parallel planes, substituting the coordinates of the given point into this equation will find that missing constant \(k\).
In our scenario, we have the point \((0, 0, 2)\), and we need to ensure that this point lies on the plane we are considering. Substituting these coordinates into \(x + y + z = k\), we end up with \(0 + 0 + 2 = k\), which simplifies to \(k=2\).
This solution step is crucial because it takes a generic equation and makes it specific to a given plane, thereby narrowing down the infinite possibilities to just one unique solution. Now, the equation \(x + y + z = 2\) is not just any plane among the parallel set; it is precisely the one that includes the point \((0, 0, 2)\). This highlights how a single point can anchor a plane's exact position in three-dimensional space, distinguishing it from all other parallel companions.
In our scenario, we have the point \((0, 0, 2)\), and we need to ensure that this point lies on the plane we are considering. Substituting these coordinates into \(x + y + z = k\), we end up with \(0 + 0 + 2 = k\), which simplifies to \(k=2\).
This solution step is crucial because it takes a generic equation and makes it specific to a given plane, thereby narrowing down the infinite possibilities to just one unique solution. Now, the equation \(x + y + z = 2\) is not just any plane among the parallel set; it is precisely the one that includes the point \((0, 0, 2)\). This highlights how a single point can anchor a plane's exact position in three-dimensional space, distinguishing it from all other parallel companions.
