Question

17\. Find the equation of the plane that contains the parallel lines $$ \left\\{\begin{array} { l } { x = - 2 + 2 t } \\ { y = 1 + 4 t \quad \text { and } } \\ { z = 2 - t } \end{array} \left\\{\begin{array}{l} x=2-2 t \\ y=3-4 t \\ z=1+t \end{array}\right.\right. $$

Short answer

The plane's equation is \(x - y + 6z = -11\).

Step-by-step solution

Write the vector form of the lines

The first line can be represented with the vector form as: \( \mathbf{r}_1(t) = \begin{pmatrix} -2 \ 1 \ 2 \end{pmatrix} + t \begin{pmatrix} 2 \ 4 \ -1 \end{pmatrix} \).The second line can be represented with the vector form as: \( \mathbf{r}_2(s) = \begin{pmatrix} 2 \ 3 \ 1 \end{pmatrix} + s \begin{pmatrix} -2 \ -4 \ 1 \end{pmatrix} \).

Determine direction vectors of the lines

For line 1, the direction vector is \( \mathbf{d}_1 = \begin{pmatrix} 2 \ 4 \ -1 \end{pmatrix} \).For line 2, the direction vector is \( \mathbf{d}_2 = \begin{pmatrix} -2 \ -4 \ 1 \end{pmatrix} \). Since the lines are parallel, both vectors are scalar multiples of each other.

Choose a point from each line

For line 1, choose the point \( P_1(-2, 1, 2) \).For line 2, choose the point \( P_2(2, 3, 1) \).

Find a normal vector to the plane

Calculate the vector \( \mathbf{v} \) connecting \( P_1 \) and \( P_2 \): \( \mathbf{v} = P_2 - P_1 = \begin{pmatrix} 2 \ 3 \ 1 \end{pmatrix} - \begin{pmatrix} -2 \ 1 \ 2 \end{pmatrix} = \begin{pmatrix} 4 \ 2 \ -1 \end{pmatrix} \).Using \( \mathbf{d}_1 \) and \( \mathbf{v} \), find a normal vector \( \mathbf{n} \) to the plane by taking their cross product:\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 4 & -1 \ 4 & 2 & -1 \end{vmatrix} = \begin{pmatrix} -2 \ 2 \ -12 \end{pmatrix} \].

Formulate the equation of the plane using a point and normal vector

Using the normal vector \( \mathbf{n} = \begin{pmatrix} -2 \ 2 \ -12 \end{pmatrix} \) and point \( P_1(-2, 1, 2) \) in the equation of the plane:\[ -2(x + 2) + 2(y - 1) - 12(z - 2) = 0 \].This simplifies to: \[ -2x + 2y - 12z = 22 \].

Simplify the equation

Divide through by -2 (or re-arrange) to further simplify the plane's equation:\[ x - y + 6z = -11 \].

Key concepts

Vector Algebra

Vector algebra is a crucial concept in understanding how we can describe and work with lines and planes in three-dimensional space. Vectors are entities that have both magnitude and direction. They allow us to easily represent geometrical figures such as lines and planes using algebraic expressions.
In this exercise, vectors are used to describe the lines. For each line, we can express it in a vector form as a combination of a point on the line and a direction vector. This representation is vital because it helps in easily understanding the position and orientation of the lines in space.
By using vector algebra, specifically adding, subtracting, and scalar multiplication of vectors, we can find direction vectors and points that lie on the lines. This makes analyzing the spatial relations between lines and planes more manageable and understandable for students.

Cross Product

The cross product is a fundamental operation in vector algebra, used to find a vector that is perpendicular to two given vectors. It is especially useful in three-dimensional space, particularly in our problem, where we need to find a plane containing parallel lines.
To determine a plane's orientation, we exploit the property of the cross product that gives us a normal vector to the plane. The normal vector is perpendicular to the direction vectors of the lines within the plane.
For example, in this exercise, the cross product of direction vector \( \mathbf{d}_1 \) and another vector \( \mathbf{v} \) (formed between two points on the lines) helps us find the normal vector. The calculated normal vector \( \mathbf{n} = \begin{pmatrix} -2 \, 2 \, -12 \end{pmatrix} \) is essential for forming the plane equation. Understanding the cross product's mechanics is crucial for solving problems involving planes and their orientation in space.

Parametric Equations

Parametric equations are a way of expressing a set of equations that represent a line or curve, with a parameter—often denoted as \( t \) or \( s \)—that varies over some interval. These equations provide an effective way to produce a vector form of a line in three-dimensional space.
In the exercise, the lines are given in parametric form, allowing us to easily identify a point on each line and the direction vector. The parametric form clearly distinguishes each coordinate's dependence on the parameter. For instance, for the first line:

• \( x = -2 + 2t \)
• \( y = 1 + 4t \)
• \( z = 2 - t \)

These expressions define the points and illustrate how a line extends through space.
This form is particularly helpful for visualizing how changes in the parameter move the point along the line, and it is a stepping stone for deriving other expressions like vector forms or Cartesian equations of lines.

Direction Vectors

Direction vectors play a key role when dealing with lines in the context of vector algebra. They define the direction in which a line extends in three-dimensional space.
For any line represented in vector or parametric form, the direction vector is the vector that multiplicates the parameter \( t \). It indicates the line's orientation—essentially, the rate of change along each axis.
In this exercise, the direction vectors are derived directly from the parametric equations of the lines. They are given by:

• For line 1: \( \mathbf{d}_1 = \begin{pmatrix} 2 \, 4 \, -1 \end{pmatrix} \)
• For line 2: \( \mathbf{d}_2 = \begin{pmatrix} -2 \, -4 \, 1 \end{pmatrix} \)

These vectors are central in identifying how the lines relate to each other. They show whether lines are parallel or intersecting. Here, the vectors are scalar multiples of each other, showing that the lines are parallel.
Understanding direction vectors is essential in analyzing how lines behave in spaces, and they are integral in forming plane equations.