Question

Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Line through (3,4,-1) and parallel to \(2 \mathbf{i}-3 \mathbf{j}+6 \mathbf{k}\)

Short answer

Parametric: \( x = 3 + 2t, \ y = 4 - 3t, \ z = -1 + 6t \). Symmetric: \( \frac{x-3}{2} = \frac{y-4}{-3} = \frac{z+1}{6} \).

Step-by-step solution

Identify Point and Direction Vector

The given point through which the line passes is \( (3, 4, -1) \). The direction vector is parallel to \(2 \mathbf{i} - 3 \mathbf{j} + 6 \mathbf{k}\). The direction vector can be written as \((2, -3, 6)\).\

Write Parametric Equations

Using the point \( (3, 4, -1) \) and direction vector \((2, -3, 6)\), the parametric equations are given by:\[ x = 3 + 2t, \ y = 4 - 3t, \ z = -1 + 6t \]\

Derive Symmetric Equations

To find the symmetric equations, solve each parametric equation for \(t\):\[ t = \frac{x-3}{2}, \ t = \frac{y-4}{-3}, \ t = \frac{z+1}{6} \]\The symmetric form is then:\[ \frac{x-3}{2} = \frac{y-4}{-3} = \frac{z+1}{6} \]\

Key concepts

Direction Vector

Understanding the direction vector is crucial when working with lines in 3D space. A direction vector tells us the 'direction' in which the line moves. It is often represented as \( \textbf{d} = (a, b, c) \), referring to the components in the x, y, and z directions respectively. It shows how much the line moves along each axis.

In our given problem, the direction vector is \(2 \textbf{i} - 3 \textbf{j} + 6 \textbf{k} \). Breaking this down, it means the line moves 2 units in the x direction, -3 units in the y direction, and 6 units in the z direction. Understanding this can help visualize the line in 3D.

So, whenever given a direction vector or need to find one, remember:

• The vector defines the slope of the line in space.
• Write it in terms of its components \( (a, b, c) \), like \( (2, -3, 6) \).
• This vector helps to form parametric and symmetric equations for the line.

Parametric Equations

Parametric equations are a way to express the coordinates of the points that make up a line using a parameter, often denoted by t. It connects the components of the direction vector and a point through which the line passes.

In the given problem, we use the point \( (3, 4, -1) \) and the direction vector \( (2, -3, 6) \) to create the parametric equations:
\[ x = 3 + 2t, \]
\[ y = 4 - 3t, \]
\[ z = -1 + 6t \]
What we are essentially saying is:

• 'x' starts at 3 and changes by 2 for every unit of 't'
• 'y' starts at 4 and decreases by 3 for every unit of 't'
• 'z' starts at -1 and increases by 6 for every unit of 't'.

This form of representing the line using parameter 't' helps us find any point on the line by choosing different 't' values.

So, remember the essentials of parametric equations:

• Use a known point and the direction vector.
• Express each coordinate as a function of the parameter 't'.
• 't' can take any real value to signify points along the whole line.

Symmetric Equations

Symmetric equations provide another way to represent a line, and they often make it clearer how the coordinates relate to each other directly. We start with the parametric equations and eliminate the parameter 't'.

From the exercise, we had:
\[ x = 3 + 2t, \]
\[ y = 4 - 3t, \]
\[ z = -1 + 6t \]
Solving each for 't', we get:

• \( t = \frac{x-3}{2} \)
• \( t = \frac{y-4}{-3} \)
• \( t = \frac{z+1}{6} \)

The symmetric equation combines these relations:
\[ \frac{x-3}{2} = \frac{y-4}{-3} = \frac{z+1}{6} \]
This format shows the relationship between x, y, and z coordinates directly without explicitly using a parameter. It can be easier to understand the proportional changes between the coordinates.

Key points for symmetric equations:

• They simplify the relationship between coordinates by removing the parameter 't'.
• The ratios in the equations show how much each coordinate changes relative to each other.
• Start from parametric equations, solve for 't', and combine them.