Question

Give the equation of the described plane in standard and general forms. Contains the point (5,7,3) and is orthogonal to the line \(\vec{\ell}(t)=\langle 4,5,6\rangle+t\langle 1,1,1\rangle\)

Short answer

The equation of the plane is \(x+y+z=15\) or \(x+y+z-15=0\).

Step-by-step solution

Understand the Problem

We need to find the equation of a plane that contains the point \( (5,7,3) \) and is orthogonal to a given line \( \vec{\ell}(t)=\langle 4,5,6\rangle+t\langle 1,1,1\rangle \). "Orthogonal" means the plane's normal vector is parallel to the direction vector of the line.

Identify the Normal Vector

The direction vector of the line is \( \langle 1, 1, 1 \rangle \). Since the plane is orthogonal to the line, this vector is the same as the normal vector \( \mathbf{n} \) of the plane.

Plug Point into Plane Equation

A plane can be described by the equation \( Ax + By + Cz = D \), and it must pass through the point \( (5, 7, 3) \). Since the normal vector is \( \langle 1, 1, 1 \rangle \), we write the equation as \( 1(x - 5) + 1(y - 7) + 1(z - 3) = 0 \).

Simplify to Standard Form

Expanding the equation from the previous step, we have \( x - 5 + y - 7 + z - 3 = 0 \), which simplifies to \( x + y + z = 15 \). This is the standard form of the plane equation.

Convert to General Form

The general form of a plane equation is \( Ax + By + Cz + D = 0 \). From our standard form \( x + y + z = 15 \), we rearrange it as \( x + y + z - 15 = 0 \).

Key concepts

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle, also known as being perpendicular to each other. Such vectors have a dot product of zero. This property originates from algebraic principles governing vector operations. In our context, the line given by \( \vec{\ell}(t)=\langle 4,5,6\rangle+t\langle 1,1,1\rangle \) has a direction vector \( \langle 1, 1, 1 \rangle \). Because the plane is orthogonal to this line, the direction vector is also the normal vector of the plane.

Normal Vector

A normal vector is a vector that is perpendicular to a plane. It serves as a crucial element in defining the equation of a plane. If you imagine a flat surface, the normal vector pokes straight out of the surface, indicating the plane's orientation in space. In the exercise, the direction vector \( \langle 1, 1, 1 \rangle \) of the line becomes the normal vector \( \mathbf{n} \) of the plane. This normal vector helps us establish the plane's equation using its components as coefficients in the plane equation.

Standard Form

The equation of a plane in standard form is given by \( Ax + By + Cz = D \), where \( A \), \( B \), and \( C \) are components of the normal vector, and \( D \) is a constant. This form is straightforward because it directly relates to how the line intersects or positions itself concerning the plane. For the current plane, substituting the normal vector \( \langle 1, 1, 1 \rangle \) and the point \( (5, 7, 3) \), the equation first appears as \( 1(x - 5) + 1(y - 7) + 1(z - 3) = 0 \). After simple expansion, this results in the standard form equation \( x + y + z = 15 \). The simplicity of this form makes it easy to recognize the plane's orientation.

General Form

The general form of a plane equation is a representation that introduces an extra constant on the right side. It is expressed as \( Ax + By + Cz + D = 0 \), adapting from the standard form. The values \( A \), \( B \), and \( C \) are taken from the normal vector, while \( D \) represents a shifted constant from \( D = 15 \) from the standard form to \( D = -15 \) for consistent algebraic balance. So, our plane becomes \( x + y + z - 15 = 0 \). This form provides flexibility for different algebraic manipulations, maintaining equivalence in describing the plane's position in space.