Problem 55 - Calculus for Biology and Medicine

Question

In Problems \(55-58\), find the parametric equation of the line in \(x-\) \(y-z\) space that goes through the indicated point in the direction of the indicated vector. $$ (1,-1,2),\left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The parametric equations of the line are: \(x = 1 + t\), \(y = -1 - 2t\), \(z = 2 + t\).

Step 1: Understand the given information

We have a point through which the line passes, \((1, -1, 2)\), and a direction vector \([1, -2, 1]\).This means that the line passes through the point (1, -1, 2) and travels in the direction specified by the vector \([1, -2, 1]\).

Step 2: Write the general formula for a parametric line

The parametric equation of a line in 3D space can be expressed as:\[(x, y, z) = (x_0, y_0, z_0) + t \cdot (a, b, c)\]where \((x_0, y_0, z_0)\) is a point on the line, which is \((1, -1, 2)\), and\((a, b, c)\) is the direction vector, which is \((1, -2, 1)\).

Step 3: Substitute the values into the parametric equation

Substitute \((x_0, y_0, z_0) = (1, -1, 2)\) and \((a, b, c) = (1, -2, 1)\) into the general formula:\[(x, y, z) = (1, -1, 2) + t \cdot (1, -2, 1)\]Expanding this to its individual components gives us three separate equations to describe the line.

Step 4: Write the parametric equations for each component

Using the expanded form, we derive the individual parametric equations:\(x = 1 + t\)\(y = -1 - 2t\)\(z = 2 + t\)Each of these equations describes the x, y, and z coordinates of any point on the line as a function of \(t\).

Key Concepts

3D Space

In our daily lives, we're used to dealing with two-dimensional spaces, like a piece of paper or a screen. However, three-dimensional (3D) space adds an extra layer of complexity, which allows us to understand and visualize things more realistically, closer to how objects exist in the real world. In 3D space, each point is defined by three coordinates:

• x-coordinate (horizontal position)
• y-coordinate (vertical position)
• z-coordinate (depth)

These coordinates form a vector that specifies a position in the space, like \(x, y, z\). When working with lines in 3D space, we need a more sophisticated way to describe their location and direction than just using the familiar slope-intercept form from 2D space.
The parametric equations of a line in 3D use these three components to describe every point on the line, which is quite different from graphing lines on a simple x-y plane.

Direction Vector

A direction vector is crucial when we want to describe the orientation of a line in 3D space. Think of it as a pointer that shows where the line is headed. This vector doesn't start from the origin but rather emerges from any point on the line. It gives us the path along which every point on the line is aligned.

• The direction vector tells us how the line moves in terms of its x, y, and z components from one point to another on the line.
• The notation for a direction vector is usually written as \(a, b, c\), where these are the changes in the x, y, and z directions, respectively.

When creating a parametric equation, the direction vector \(1, -2, 1\) acts as the multiplier for the parameter \(t\). This affects how much we traverse through the line from the starting point for each unit increment of \(t\). It’s like saying, "for every 1 step in the x direction, move -2 in the y direction and 1 in the z direction."

Point through which line passes

To define a particular line in 3D space, we need to know a specific point that the line crosses. This point, usually denoted as \(x_0, y_0, z_0\), serves as the starting point in the parametric representation of the line. For our exercise, the point \(1, -1, 2\) is given.

• This point is significant because every point \(x, y, z\) on the line can be traced back to it using the direction vector.
• By setting \(t = 0\), the parameter equals zero, and we return to this initial point on the 3D line.

In mathematical terms, this point ensures that we have a fixed reference from which the line is generated. The parametric equations then use this point and the direction vector to give the coordinates of any point on the line as the parameter \(t\) varies. By effectively combining this point with the direction vector, we're able to describe the line's trajectory comprehensively.