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Chapter 11: Problem 15

Find the parametric equations of the line through \((5,-3,4)\) that intersects the \(z\) -axis at a right angle.

Short Answer

Expert verified
Parametric equations: \(x = 5 + at, y = -3 + bt, z = 4\).

Step by step solution

01

Understand the Problem

We need to find parametric equations for a line that passes through the point \((5, -3, 4)\) and intersects the z-axis at a right angle.
02

Determine the Direction Vector

A line perpendicular to the z-axis will have a direction vector of the form \((a, b, 0)\) because the line must be parallel to the xy-plane.
03

Write the Parametric Equations

The parametric equations for a line through the point \((5, -3, 4)\) with direction vector \((a, b, 0)\) will be:\[x = 5 + at,\quad y = -3 + bt,\quad z = 4\]
04

Ensure Z-Axis Perpendicularity

Since the line must intersect the z-axis perpendicularly, the z-component of the direction vector is 0, which is already satisfied with direction vector \((a, b, 0)\).
05

Solution Verification

The line intersects the z-axis when \(x = 0\) and \(y = 0\), since we need the line to pass through the z-axis. However, it is sufficient that the direction vector only affects x and y to ensure perpendicularity to the z-axis, as z remains constant. Thus, our set of equations is valid.

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Key Concepts

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

Perpendicular Lines
When discussing lines in three-dimensional space, two lines are perpendicular if they meet at a right angle. A line that is perpendicular to the z-axis specifically has to lie entirely within a plane parallel to the xy-plane. This is because the z-axis is vertical, so any line that crosses it perpendicularly does so without moving up or down the z-values. The essence of ensuring a perpendicular intersection with the z-axis involves understanding that the only varying components of the line's direction would be along the x and y axes, leaving the z component fixed.
Direction Vector
The direction vector of a line is crucial as it defines the line's pathway through space. In our problem, we determined our direction vector to have the form \((a, b, 0)\). This choice results because a direction vector \((a, b, 0)\) ensures that the line runs parallel to the xy-plane and thus remains perpendicular to the z-axis.
  • The x and y components \(a\) and \(b\) allow the line to move through different x and y coordinates.
  • The z component being zero ensures perpendicularity, as there is no z direction displacement.
A well-chosen direction vector can simplify computations and clearly illustrate the line's orientation in 3D space.
Z-Axis Intersection
For the line to intersect the z-axis, we usually think of it crossing the plane where the x and y coordinates are zero. However, for our specific problem, this constraint doesn't necessarily apply strictly; rather, we need the line's orientation to remain consistently aligned so it touches the z-axis at one fixed point along that axis. Consequently, the variation in x and y as defined by the parametric equations allows for looking at how these interact rather than directly demanding \(x = 0\) and \(y = 0\).
  • The critical focus is on how the line's z value remains constant, thereby automatically ensuring an intersection with the z-axis at one point along its line.
  • The direction is exclusively along the x and y path, which effectively addresses where the line covers as \(t\) changes.
Parametric Form in 3D
Parametric equations in 3D define a line using separate expressions for each coordinate as a function of a common parameter. Here, we use \(t\), with equations:
  • \(x = 5 + at\)
  • \(y = -3 + bt\)
  • \(z = 4\)
This method allows for capturing all points along the line's path by letting \(t\) vary. To reiterate:
  • The starting point of the line is \((5, -3, 4)\), and \(t = 0\) will plug directly into this origin.
  • As \(t\) shifts, it records how far we've moved from this initial point, controlled by our chosen direction vector.
  • The equations provide the line's representation in clear numerical terms that describe both location and direction in 3D space.
Using parametric form in space is powerful as it succinctly combines multiple spatial dimensions into one cohesive set of equations that communicate movement and position.

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Most popular questions from this chapter

The medians of a triangle meet at a point \(P\) (the centroid by Problem 30 of Section 6.6) that is two-thirds of the way from a vertex to the midpoint of the opposite edge. Show that \(P\) is the head of the position vector \((\mathbf{a}+\mathbf{b}+\mathbf{c}) / 3\), where \(\mathbf{a}, \mathbf{b}\), and \(\mathbf{c}\) are the position vectors of the vertices, and use this to find \(P\) if the vertices are \((2,6,5),(4,-1,2)\), and \((6,1,2)\).

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