Question

Give the equation of the described plane in standard and general forms. Contains the parallel lines \(\vec{\ell}_{1}(t)=\langle 1,1,1\rangle+t\langle 4,1,3\rangle\) and $$ \vec{\ell}_{2}(t)=\langle 4,4,4\rangle+t\langle 4,1,3\rangle $$

Short answer

The plane in standard form is \(-3y = -3\), or \(y = 1\) in general form.

Step-by-step solution

Identify the direction vector of the plane

Since the plane contains two parallel lines \( \vec{\ell}_{1} \) and \( \vec{\ell}_{2} \), both lines have the same direction vector \( \langle 4,1,3 \rangle \). This vector will also be a direction vector of the plane.

Find a point on each line

Select a point from each line: \( \vec{A} = \langle 1,1,1 \rangle\) from \( \vec{\ell}_{1} \) and \( \vec{B} = \langle 4,4,4 \rangle\) from \( \vec{\ell}_{2} \).

Find a vector between the two points

Compute the vector between the points \( \vec{A} = \langle 1,1,1 \rangle \) and \( \vec{B} = \langle 4,4,4 \rangle \) as \( \vec{AB} = \langle 4,4,4 \rangle - \langle 1,1,1 \rangle = \langle 3,3,3 \rangle \).

Determine the normal vector to the plane

The normal vector \( \vec{n} \) to the plane is orthogonal to both direction vectors \( \langle 4,1,3 \rangle \) and \( \langle 3,3,3 \rangle \). Compute the cross product: \( \vec{n} = \langle 4,1,3 \rangle \times \langle 3,3,3 \rangle \).

Calculate the cross product

Using the formula for the cross product, get \( \vec{n} = \langle 1(3) - 3(1), 3(3) - 3(4), 4(1) - 1(4) \rangle = \langle 0, -3, 0 \rangle \).

Write the plane equation in standard form

Since the normal vector \( \vec{n} = \langle 0, -3, 0 \rangle \), substitute into the plane equation \( ax + by + cz = d \). Choose point \( \langle 1,1,1 \rangle \), resulting in \( 0(1) - 3(1) + 0(1) = d \), so \( d = -3 \). Thus, the standard form is \( -3y = -3 \).

Simplify to obtain the plane in general form

Rewriting \( -3y = -3 \), we get \( y = 1 \) as the general form of the plane equation.

Key concepts

Vectors

Vectors are fundamental in understanding various mathematical and physics-related concepts. Essentially, a vector is an object that has both magnitude and direction. In simpler terms, it is like an arrow pointing from one location to another in space. This arrow, or vector, can be represented in a coordinate system using a set of numerical components. For instance, the vector \( \langle 4,1,3 \rangle \) indicates that the vector moves 4 units along the x-axis, 1 unit along the y-axis, and 3 units along the z-axis.

When dealing with vectors, several operations can be performed, including addition, scalar multiplication, and vector multiplication. Vectors are crucial in fields such as engineering, physics, and computer graphics, as they can depict forces, velocities, and more. Understanding vectors allows you to navigate through many spatial problems efficiently.

• Components: The values in the vector, showing magnitude along each axis.
• Direction: The way in which the vector points.
• Magnitude: The length of the vector, calculated by the square root of the sum of the squares of its components.

Planes

Planes are two-dimensional surfaces that extend infinitely in all directions within a three-dimensional space. A plane can be thought of as a flat sheet that has no thickness, perfectly even everywhere. Mathematically, a plane is defined using a point and a vector. The vector is perpendicular to the plane and is known as the normal vector.

For example, in our original problem, we derive a plane's equation using points and vectors. A single plane can be described in standard form with an equation \( ax + by + cz = d \), where \( a \), \( b \), and \( c \) are the components of the normal vector. The constant \( d \) defines the plane's distance from the origin.
Many real-world problems, like finding the angle between surfaces or modeling the wings of an airplane, involve working with planes.

• Normal vector: A vector perpendicular to the plane.
• Standard form: The algebraic representation depicting the plane.
• General form: A simplified version of the plane’s equation.

Cross Product

The cross product is a vector multiplication operation used widely in three-dimensional space. It involves two vectors and results in a third vector that is perpendicular to the plane formed by the first two vectors. This operation is essential in finding normal vectors to surfaces such as planes.

In the problem at hand, the cross product \( \langle 4,1,3 \rangle \times \langle 3,3,3 \rangle \) yields \( \langle 0, -3, 0 \rangle \). This resulting vector is indeed perpendicular to both initial vectors, acting as the normal vector defining our plane.
The cross product is denoted by \( \times \) and can be computed using the determinant of a matrix formed by the component vectors.

• Resultant vector: Always orthogonal to the original pair.
• Utility: Used to find perpendicular directions and calculate areas.
• Direction: Determined by the right-hand rule.

Direction Vector

A direction vector is a vector that provides information about the direction that a line travels. It is an integral part of vector calculus, especially when defining lines and planes. For lines, the direction vector determines how the line moves through space, and for planes, it can define their orientation through parallel lines.

In the exercise, the direction vector \( \langle 4,1,3 \rangle \) is used to denote the direction in which both lines \( \vec{\ell}_{1} \) and \( \vec{\ell}_{2} \) extend. Since these lines are parallel, they share the same direction vector. Thus, this vector also helps to characterize the orientation of the entire plane within three-dimensional space.
Understanding direction vectors is crucial for solving problems that involve path tracing or determining parallelism between lines or curves.

• Describes the path or direction.
• Essential for understanding geometry in space.
• Can be used to solve parallel line equations.