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Find an equation of the plane that passes through the point \(P_{0}\) with a normal vector \(\mathbf{n}\). $$P_{0}(1,0,-3) ; \mathbf{n}=\langle 1,-1,2\rangle$$

Short Answer

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Question: Find the equation of the plane that passes through the point \(P_{0}(1,0,-3)\) with a normal vector \(\mathbf{n}=\langle 1,-1,2\rangle\). Answer: The equation of the plane is given by \(x - y + 2z = -5\).

Step by step solution

01

Recalling the general form of a plane equation

Begin with the general form of the equation of a plane: $$Ax+By+Cz=D$$
02

Substituting the normal vector

Substitute the normal vector \(\mathbf{n}=\langle 1,-1,2\rangle\) into the general form of the equation of a plane: $$1x - 1y + 2z = D$$
03

Substituting the point P_0

Substitute the given point \(P_{0} = (1,0,-3)\) into the equation: $$1(1) - 1(0) + 2(-3) = D$$
04

Solving for D

After substituting the point and normal vector, simplify the equation to find the value of \(D\) to get: $$1 - 0 - 6 = D$$ $$D = -5$$
05

Writing the final equation

Now that we have found the value of \(D\), we can plug it back into the general equation form to get the equation of the plane as: $$x - y + 2z = -5$$

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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, a normal vector plays a crucial role in defining planes. Think of a normal vector as an arrow pointing directly perpendicular to the surface of a flat plane. If you imagine a sheet of paper, the normal vector would point straight out of it. Knowing the normal vector is essential because it helps in determining how the plane is oriented in space.

The given normal vector in this exercise is \(\mathbf{n} = \langle 1, -1, 2 \rangle\). This vector gives us important coefficients for the equation of the plane. In simple terms, these components \(1\), \(-1\), and \(2\) tell us how steeply the plane rises or falls in each direction of the 3D space.

In the realm of 3D geometry, these three numbers connect directly to the x, y, and z axes, showing how changes in these directions affect the plane's position.
Equation of a Plane
The equation of a plane in three-dimensional space is expressed in the standard form \(Ax + By + Cz = D\). Each letter represents a specific aspect of the plane:
  • \(A\), \(B\), and \(C\) are the coefficients derived from the normal vector.
  • \(D\) is a constant determined by substituting a specific point on the plane into the equation.
For our exercise, we used a normal vector \(\langle 1, -1, 2 \rangle\), leading to the equation \(1x - 1y + 2z = D\).

By plugging in the coordinates of the point \(P_{0} = (1,0,-3)\), we calculate \(D\) as follows:
\(1 \times 1 - 1 \times 0 + 2 \times (-3) = D\), which simplifies to \(1 - 0 - 6 = -5\).
Thus, the final equation of the plane becomes \(x - y + 2z = -5\). This equation now fully represents the flat surface extending infinitely in all directions across three-dimensional space, defined by the given point and normal vector.
3D Geometry
Three-dimensional (3D) geometry explores the properties and relationships of geometric figures in space with height, width, and depth. Unlike the flat, two-dimensional planes you might be used to in more basic geometry, 3D geometry opens up a world of figures like spheres, cubes, and, importantly, planes that extend in all three directions.

Understanding coordinate systems is vital in 3D geometry. We typically use a Cartesian coordinate system with three axes: x, y, and z. Each point in 3D space is defined by these three coordinates. In our exercise, the point \(P_{0} = (1,0,-3)\) provides a specific location for the plane. The coordinates tell how far along the x, y, and z axes that point is situated.

Seeing how planes and lines interact in this space becomes crucial, especially when analyzing intersections, angles, and distances. Mastery of these interactions is critical in fields like engineering, computer graphics, and physics, where understanding complex shapes and spaces is essential.

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