Problem 22 - Advanced Engineering Mathematics

Question

In Problems, find parametric and symmetric equations for the line through the given point parallel to the given vector. $$ (0,-3,10), \mathbf{a}=\langle 12,-5,-6\rangle $$

Step-by-Step Solution

Answer

Parametric: \(x = 12t, y = -3 - 5t, z = 10 - 6t\); Symmetric: \(\frac{x}{12} = \frac{y + 3}{-5} = \frac{z - 10}{-6}\).

Step 1: Identify the Components

We need to write equations for a line that passes through the point given as \(0, -3, 10\) and is parallel to the vector \mathbf{a} = \langle 12, -5, -6 \rangle\. The point will give us the position vector, and the components of the vector \mathbf{a}\ will be the direction ratios.

Step 2: Determine Parametric Equations

The parametric equations for a line require a point and a direction. From the point \(0, -3, 10\), our parametric equations can be written by adding scalar multiples of \(t\) (a parameter) with the direction vector \langle 12, -5, -6 \rangle\. Thus, the parametric equations are: \begin{align*} x &= 12t, \ y &= -3 - 5t, \ z &= 10 - 6t. \end{align*} \

Step 3: Derive Symmetric Equations

The symmetric equations eliminate the parameter \(t\) using the parametric equations. From the equations \(x = 12t\), \(y = -3 - 5t\), and \(z = 10 - 6t\), we solve for \(t\): \begin{align*} t &= \frac{x}{12}, \ t &= \frac{y + 3}{-5}, \ t &= \frac{z - 10}{-6}. \end{align*} By setting these equal, the symmetric equations become: \frac{x}{12} = \frac{y + 3}{-5} = \frac{z - 10}{-6}.\

Key Concepts

Symmetric Equations

Symmetric equations are a form of linear equations that describe a line in three-dimensional space without using a parameter like "t" in parametric equations. They are particularly useful because they provide a single equation format showing the relationship between three variables (x, y, and z) relative to each other. In essence, they are derived from the parametric form by solving for the parameter and equating the expressions.

For instance, if you have parametric equations given by \(x = 12t\), \(y = -3 - 5t\), and \(z = 10 - 6t\), you solve these equations for "t":

• \(t = \frac{x}{12}\)
• \(t = \frac{y + 3}{-5}\)
• \(t = \frac{z - 10}{-6}\)

Once you have these expressions, set them equal to create the symmetric equation:

\[ \frac{x}{12} = \frac{y + 3}{-5} = \frac{z - 10}{-6} \]This provides a consolidated view and is particularly valuable for solving problems or analyzing a line's properties in a simpler form.

Direction Vector

The direction vector of a line in space is a vector that gives the direction along which the line points. It plays a crucial role in constructing both parametric and symmetric equations for lines. For any line, the direction vector is a key component in defining how the line extends infinitely in both directions.

A direction vector can be written in the vector form \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \). In this particular exercise, the direction vector \( \mathbf{a} \) is given as \(\langle 12, -5, -6 \rangle\).

Using the direction vector, we express how each coordinate (x, y, and z) changes with respect to the parameter \( t \) in parametric equations:

• \(x = a_1 t\)
• \(y = a_2 t\)
• \(z = a_3 t\)

This information is instrumental for parametrizing the line and eventually converting those parameters into symmetric equations. The direction vector tells you where the line heads in space, considering both orientation and magnitude.

Position Vector

The position vector is a vector that indicates a specific point on a line in space. It is derived from a point given on the line, such as when problems provide coordinates through which the line passes.

In the discussed exercise, the provided point is \((0, -3, 10)\). This point is used to construct the position vector \( \mathbf{r}_0 \), often written as \( \langle x_0, y_0, z_0 \rangle \), leading to the vector \( \langle 0, -3, 10 \rangle \) in our case.

The position vector acts as the starting point for the parametric equations of the line:

• \(x = x_0 + a_1 t\)
• \(y = y_0 + a_2 t\)
• \(z = z_0 + a_3 t\)

This is crucial as it dictates precisely where the line is situated in space initially before extending infinitely in the direction specified by the direction vector. In combination with the direction vector, it fully defines the line's position and direction.