Problem 2 - University Calculus: Early Transcendentals

Question

Find parametric equations for the lines. The line through \(P(1,2,-1)\) and \(Q(-1,0,1)\)

Step-by-Step Solution

Answer

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

Step 1: Find the Direction Vector

To find the direction vector \( \mathbf{v} \) of the line, subtract the coordinates of point \( P \) from those of \( Q \). This gives: \( \mathbf{v} = Q - P = (-1 - 1, 0 - 2, 1 - (-1)) =(-2, -2, 2) \).

Step 2: Use a Point on the Line

You can use either point \( P(1,2,-1) \) or \( Q(-1,0,1) \) to write the parametric equations. We will use point \( P \).

Step 3: Write the Parametric Equations

The parametric equations of a line can be written as:\[\begin{align*}x &= x_0 + at \y &= y_0 + bt \z &= z_0 + ct\end{align*}\]where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is the direction vector. Substituting point \( P(1,2,-1) \) and direction vector \( \mathbf{v} = (-2, -2, 2) \), we get:\[\begin{align*}x &= 1 - 2t \y &= 2 - 2t \z &= -1 + 2t\end{align*}\]

Key Concepts

Direction vector

A direction vector plays a crucial role in defining the line in space. It helps us understand the direction in which the line extends. When we have two points on a line, we can find the direction vector by subtracting the coordinates of one point from another.

For instance, given two points, \( P(1, 2, -1) \) and \( Q(-1, 0, 1) \), we can find the direction vector \( \mathbf{v} \) by calculating \( Q - P \). The operation is done component-wise:

• \( x \)-component: \( -1 - 1 = -2 \)
• \( y \)-component: \( 0 - 2 = -2 \)
• \( z \)-component: \( 1 - (-1) = 2 \)

This gives us a direction vector \( \mathbf{v} = (-2, -2, 2) \). Having the direction vector, we can now describe how the line aligns and spreads in a three-dimensional space.

Points on a line

When working with parametric equations, knowing at least one point on the line is essential. The parametric equations use this point to tell us the starting position of the line in a space, blending together with a direction vector.

In our exercise, points \( P(1, 2, -1) \) and \( Q(-1, 0, 1) \) are both located on the line we are investigating. Either of these points can be used interchangeably to formulate the parametric equations. This flexibility offers us options, depending on which computations might seem more straightforward or intuitive.

While the choice of point doesn't change the direction of the line, it does affect the specific equations we get, as they describe the path starting from that particular location. Thus, picking the point can simplify calculations depending on our preference.

Parametric form of a line

The parametric form of a line is a very helpful way of expressing lines in three-dimensional space. It uses a combination of a point on the line and a direction vector to describe every possible point along the line. This form is highly useful in solving problems involving lines crossing, intersection points, or even parallelism.

To write a line in parametric form, start by choosing a point \((x_0, y_0, z_0)\) from the line and a direction vector \((a, b, c)\). The equations are:

• \(x = x_0 + at\)
• \(y = y_0 + bt\)
• \(z = z_0 + ct\)

Plugging our examples into this format using point \(P(1, 2, -1)\) and direction vector \((-2, -2, 2)\), we generate:

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

In these equations, \( t \) is a parameter that can take any real number value, determining each point's contribution from the start point along the line’s direction. The line's path is effectively traced out as \( t \) varies.