Question

Give the equation of the described plane in standard and general forms. Contains the point (1,2,3) and is parallel to the plane \(x=5 .\)

Short answer

The plane is \(x = 1\) in both standard and general forms.

Step-by-step solution

Understand the problem

We need to find the equation of a plane that passes through a given point (1, 2, 3) and is parallel to another plane with the equation \(x = 5\). We need to express this plane in both standard and general forms.

Identify the normal vector

A plane with the equation \(x = 5\) has the normal vector \((1,0,0)\) because the equation is independent of \(y\) and \(z\). Consequently, a plane parallel to \(x = 5\) will have the same normal vector.

Write the standard form of the equation

In standard form, a plane with normal vector \((A,B,C)\) passing through a point \((x_0, y_0, z_0)\) is given by: \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\). With \((A,B,C) = (1,0,0)\) and point \((1,2,3)\), the equation becomes: \(1(x - 1) + 0(y - 2) + 0(z - 3) = 0\), simplifying to \(x - 1 = 0\).

Simplify and state the standard form

The standard form of the plane is \(x = 1\).

Convert to general form

The general form of a plane equation is \(Ax + By + Cz = D\). For our plane, since \(x = 1\), it can be expressed as \(1x + 0y + 0z = 1\), which is \(x = 1\). In generalized format, \(x - 1 = 0\) can be rewritten as \(x - 1 = 0\).

Verify the forms

Both standard form \(x = 1\) and general form \(x - 1 = 0\) describe the same geometrical plane that is parallel to \(x = 5\) and passes through \( (1, 2, 3) \).

Key concepts

Normal Vector

In geometry, a plane can be uniquely identified by its normal vector. A normal vector is a vector that is perpendicular to the plane. It's essential in defining the orientation of the plane in a 3D space. For a plane given by the equation \( Ax + By + Cz = D \), the normal vector is \((A, B, C)\). This vector does not lie on the plane, but rather extends out of it, indicating how the plane is tilted in space.

• The normal vector helps us understand which planes are parallel. Two planes are parallel if they have the same normal vector or if one is a scalar multiple of the other.
• In our exercise, the plane \( x = 5 \) has the normal vector \( (1, 0, 0) \), meaning it only varies in the direction of the x-axis and is parallel to all planes like it.

Understanding the normal vector is crucial for deducing various properties of the plane, such as angles between planes, intersections, and perpendicularity.

Standard Form of Plane

The standard form of a plane is particularly straightforward and allows us to determine the plane's equation using a point on the plane and its normal vector. The standard form is given by:
\[ A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \]In this equation, \((x_0, y_0, z_0)\) is any specific point through which the plane passes. The coefficients \((A, B, C)\) are the components of the normal vector.

• This form is very useful because it highlights the role of the normal vector in defining the plane.
• For the plane in our problem, the equation reduces to \(1(x - 1) + 0(y - 2) + 0(z - 3) = 0\), simplifying to \(x - 1 = 0\), or \(x = 1\).

This simplification stems from the fact that the plane’s normal vector is \((1, 0, 0)\), indicating that movement along y or z does not change the nature of this plane.

General Form of Plane

While the standard form emphasizes the plane's relationship to a specific point, the general form of a plane simplifies the expression and is straightforwardly a basic representation:
\[ Ax + By + Cz = D \]Here, \(A\), \(B\), and \(C\) remain the components of the normal vector, and \(D\) is the constant term calculated based on the given point the plane passes through.

• The general form is useful for algebraic manipulation and helps in identifying parallel planes and in some cases, intersecting points.
• In the given problem, the general form is expressed simply as \(x - 1 = 0\), or equivalently \(x = 1\), which reflects the plane’s dependency only on the x-coordinate.

This form confirms much about the geometry: consistency with the standard form, and verifying that the plane meets the initial condition of being parallel to \(x = 5\).

Parallel Planes

Two planes are considered parallel if their normal vectors are the same or scalar multiples of each other. Parallel planes never intersect and maintain a constant distance apart from each other. This property is evident when two plane equations share identical or proportionate coefficients corresponding to their \(x\), \(y\), and \(z\) components.

• In our problem, since both the given plane \(x = 1\) and the reference plane \(x = 5\) share the same normal vector \((1, 0, 0)\), they are parallel.
• The parallel nature of these planes is confirmed by the absence of terms involving \(y\) or \(z\), showing their complete independence from these axes.

Understanding parallel planes is fundamental in spatial geometry as it allows predicting behavior of surfaces in space, informing on constructs like slabs or layers in design and architecture. The characteristics of these planes ensure stability and uniformity when required.