Question

In Exercises 5-14, write the vector, parametric and symmetric equations of the lines described. Passes through \(P=(2,-4,1),\) parallel to \(\vec{d}=\langle 9,2,5\rangle\)

Short answer

Vector: \( \vec{r} = \langle 2, -4, 1 \rangle + t\langle 9, 2, 5 \rangle \). Parametric: \( x = 2 + 9t, y = -4 + 2t, z = 1 + 5t \). Symmetric: \( \frac{x - 2}{9} = \frac{y + 4}{2} = \frac{z - 1}{5} \)."}

Step-by-step solution

Identify the Given Information

We are given a point \( P = (2, -4, 1) \) that lies on the line and a direction vector \( \vec{d} = \langle 9, 2, 5 \rangle \) that indicates the direction of the line.

Write the Vector Equation

The vector equation of a line through a point \( P_0 = (x_0, y_0, z_0) \) and parallel to vector \( \vec{d} = \langle a, b, c \rangle \) is given by \( \vec{r} = \vec{r_0} + t\vec{d} \).Here, \( \vec{r_0} = \langle 2, -4, 1 \rangle \) and \( \vec{d} = \langle 9, 2, 5 \rangle \), so the vector equation is\[ \vec{r} = \langle 2, -4, 1 \rangle + t \langle 9, 2, 5 \rangle. \]

Write the Parametric Equations

The parametric equations for the line can be derived from the vector equation \( \vec{r} = \langle 2 + 9t, -4 + 2t, 1 + 5t \rangle \) by equating components:- \( x = 2 + 9t \) - \( y = -4 + 2t \) - \( z = 1 + 5t \)

Write the Symmetric Equations

The symmetric equations are derived by eliminating the parameter \( t \) from the parametric equations:- From \( x = 2 + 9t \), we have \( t = \frac{x - 2}{9} \).- From \( y = -4 + 2t \), we have \( t = \frac{y + 4}{2} \).- From \( z = 1 + 5t \), we have \( t = \frac{z - 1}{5} \).Set these equal to express the symmetric form:\[ \frac{x - 2}{9} = \frac{y + 4}{2} = \frac{z - 1}{5}. \]

Key concepts

Vector Equation of a Line

In three-dimensional space, the vector equation of a line involves position and direction components. Imagine a straight path starting at a point and heading in a direction—this is essentially what a line in 3D is about. To express this line mathematically, we use a vector equation. Given a point on the line, \( P_0 = (x_0, y_0, z_0) \), and a direction vector, \( \vec{d} = \langle a, b, c \rangle \), the vector equation is written as: \[ \vec{r} = \vec{r_0} + t\vec{d} \] where \( \vec{r_0} = \langle x_0, y_0, z_0 \rangle \) and \( t \) is a scalar parameter.
This parameter \( t \) controls how far you move along the direction vector, essentially dragging the point along the infinite line.
Thus, using the provided example its equation becomes: \[ \vec{r} = \langle 2, -4, 1 \rangle + t \langle 9, 2, 5 \rangle \]. This clearly shows how the line passes through the point (2, -4, 1) and moves in the direction \( \langle 9, 2, 5 \rangle \). Through this method, you can identify any point on the line, helping to understand the line's form in space.

Parametric Equations

To get a deeper insight into the line's path, we convert the vector equation into parametric equations. Parametric equations break down the line into three separate equations, each representing a coordinate axis: x, y, and z.
The conversion relies on the vector equation. For our exercise: - Start with \( \vec{r} = \langle 2, -4, 1 \rangle + t \langle 9, 2, 5 \rangle \)- The resulting equations are:

• \( x = 2 + 9t \)
• \( y = -4 + 2t \)
• \( z = 1 + 5t \)

Each parameter \( t \) in these equations specifies the position of corresponding x, y, and z along the line.
By changing \( t \), you can move to any desired point on the line, helping you visualize the line's trajectory through 3D space. These equations offer a clear method to pinpoint exact coordinates along the line, fence off regions of interest, or simulate motion.

Symmetric Equations

Moving from parametric to symmetric equations simplifies the representation of a line by removing the parameter \( t \). The goal is to create a single equation, making interpretation easier without explicitly tracking \( t \).
Starting from the parametric equations:

• \( x = 2 + 9t \)
• \( y = -4 + 2t \)
• \( z = 1 + 5t \)

You isolate \( t \) in each:

• From \( x = 2 + 9t \), \( t = \frac{x - 2}{9} \)
• From \( y = -4 + 2t \), \( t = \frac{y + 4}{2} \)
• From \( z = 1 + 5t \), \( t = \frac{z - 1}{5} \)

Set these expressions equal: \[ \frac{x - 2}{9} = \frac{y + 4}{2} = \frac{z - 1}{5} \]This symmetric form succinctly presents the line, highlighting its uniform nature across all axes.
Symmetric equations are particularly useful when you need to view the line in an equation form that doesn't explicitly depend on time or parameters, making it excellent for comparison with planes or other lines.