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Find a symmetric form for the line througl \(P_{\text {and }} P_{2}\) $$ P_{1}(-3,1,-1), \quad P_{2}(7,11,-8) $$

Short Answer

Expert verified
The symmetric form of the line is \( \frac{x + 3}{10} = \frac{y - 1}{10} = \frac{z + 1}{-7} \).

Step by step solution

01

Find Direction Vector

To form the direction of the line, we first need to calculate the direction vector from point \( P_1(x_1, y_1, z_1) = (-3, 1, -1) \) to point \( P_2(x_2, y_2, z_2) = (7, 11, -8) \). The direction vector \( \mathbf{d} \) is determined by subtracting the coordinates of \( P_1 \) from \( P_2 \): \[ \mathbf{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (7 - (-3), 11 - 1, -8 - (-1)) = (10, 10, -7). \]
02

Write Parameterization of the Line

Using \( P_1 \) as a reference point, the line can be parameterized by using its parametric equations: \[ x = -3 + 10t, \] \[ y = 1 + 10t, \] \[ z = -1 - 7t, \] where \( t \) is a parameter.
03

Derive Symmetric Form from Parametric Equations

To derive the symmetric form, we express each parametric equation with \( t \):\[ \frac{x + 3}{10} = t, \] \[ \frac{y - 1}{10} = t, \] \[ \frac{z + 1}{-7} = t. \] Thus, the symmetric form of the line is obtained by setting all expressions equal to \( t \) and getting: \[ \frac{x + 3}{10} = \frac{y - 1}{10} = \frac{z + 1}{-7}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector
A direction vector is fundamental when determining the orientation of a line in geometry. It is a vector that points along the line, giving the direction the line follows. In three-dimensional geometry, a direction vector is necessary to form equations of lines, such as parametric and symmetric equations.

Let's explore how we calculate a direction vector. Consider two points, which I'm going to call \( P_1(x_1, y_1, z_1) \) and \( P_2(x_2, y_2, z_2) \). You form the direction vector by subtracting the coordinates of \( P_1 \) from those of \( P_2 \). This gives us a new vector \( \mathbf{d} \).

More precisely, for our example, we calculated the direction vector as:
- \( x \)-component: \( x_2 - x_1 \)
- \( y \)-component: \( y_2 - y_1 \)
- \( z \)-component: \( z_2 - z_1 \)

Thus, the direction vector \( \mathbf{d} \) becomes \((10, 10, -7)\). With this vector, you are able to define the path the line takes from one point to another.
Parametric Equations
Parametric equations take advantage of a direction vector to represent a line in three dimensions. They give us a powerful tool to express the location of any point on the line with just one parameter. This parameter, often denoted as \( t \), varies over the real numbers.

Forming parametric equations requires you to use a point on the line and the direction vector.
- Start with a point, such as \( P_1(-3, 1, -1) \).
- Apply the direction vector \((10, 10, -7)\).

Each component of the line's location is represented as:
- \( x = x_1 + d_x\cdot t \) where \( d_x \) is the \( x \)-component of the direction vector.
- \( y = y_1 + d_y\cdot t \) where \( d_y \) is the \( y \)-component.
- \( z = z_1 + d_z\cdot t \) where \( d_z \) is the \( z \)-component.

In our case, these yield the parametric equations:
- \( x = -3 + 10t \)
- \( y = 1 + 10t \)
- \( z = -1 - 7t \)

Now just by changing \( t \), you can find any point along the line in three-dimensional space.
Three-Dimensional Geometry
Three-dimensional geometry introduces complexity by adding a third axis, the \( z \)-axis, complementing the \( x \) and \( y \) axes from 2D geometry. This additional axis enables the modeling of realistic spatial problems, making it crucial in various fields like engineering and physics.

Lines in three-dimensional geometry can be expressed using parametric and symmetric equations. While parametric equations are great for identifying specific points, symmetric equations offer a holistic perspective on the line itself.

The symmetric equation is derived from the parametric form. By equating all expressions for \( t \) from the parametric equations:
- \( \frac{x + 3}{10} = t \)
- \( \frac{y - 1}{10} = t \)
- \( \frac{z + 1}{-7} = t \)

Combining these to eliminate \( t \) gives the symmetric equation:
- \( \frac{x + 3}{10} = \frac{y - 1}{10} = \frac{z + 1}{-7} \)

This form clearly shows the relationships between coordinates of points along the line. Understanding these equations enhances our ability to work with spatial relationships in three-dimensional spaces.

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