/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Determine whether the line and p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 16

Determine whether the line and plane are parallel, perpendicular, or neither. $$ \begin{array}{l}{\text { (a) } x=3-t, \quad y=2+t, \quad z=1-3 t} \\ {\quad 2 x+2 y-5=0} \\ {\text { (b) } x=1-2 t, \quad y=t, \quad z=-t} \\ {6 x-3 y+3 z=1} \\ {\text { (c) } x=t, \quad y=1-t, z=2+t} \\ {x+y+z=1}\end{array} $$

Short Answer

Expert verified
(a) Line is parallel to the plane. (b) Line is perpendicular to the plane. (c) Line is neither parallel nor perpendicular to the plane.

Step by step solution

01

Identify Direction Vector of the Line - (a)

To find out if the line given by the parametric equations \( x = 3 - t, y = 2 + t, z = 1 - 3t \) is parallel or perpendicular to the plane, we first extract the direction vector of the line. The coefficients of \( t \) give us the direction vector \( \mathbf{d} = (-1, 1, -3) \).
02

Identify Normal Vector of the Plane - (a)

A plane given in the form \( ax + by + cz = d \) has a normal vector \( \mathbf{n} = (a, b, c) \). For the equation \( 2x + 2y - 5 = 0 \), the normal vector is \( \mathbf{n} = (2, 2, 0) \).
03

Check for Parallelism - (a)

The line and plane are parallel if the direction vector of the line is perpendicular to the normal vector of the plane, which occurs when their dot product is zero. Calculate the dot product: \( (-1)*2 + 1*2 + (-3)*0 = -2 + 2 = 0 \). Since the dot product is zero, the line is parallel to the plane.
04

Identify Direction Vector of the Line - (b)

For the line given by \( x = 1 - 2t, y = t, z = -t \), the direction vector is \( \mathbf{d} = (-2, 1, -1) \).
05

Identify Normal Vector of the Plane - (b)

For the equation \( 6x - 3y + 3z = 1 \), the normal vector is \( \mathbf{n} = (6, -3, 3) \).
06

Check for Perpendicularity - (b)

The line and plane are perpendicular if the direction vector of the line and the normal vector of the plane are parallel, i.e., they are scalar multiples of each other. Compare direction vector \( \mathbf{d} = (-2, 1, -1) \) and normal vector \( \mathbf{n} = (6, -3, 3) \). Clearly, \( (6, -3, 3) = -3(-2, 1, -1) \), confirming they are parallel, hence the line is perpendicular to the plane.
07

Identify Direction Vector of the Line - (c)

For \( x = t, y = 1 - t, z = 2 + t \), the direction vector is \( \mathbf{d} = (1, -1, 1) \).
08

Identify Normal Vector of the Plane - (c)

For \( x + y + z = 1 \), the normal vector is \( \mathbf{n} = (1, 1, 1) \).
09

Check for Neither Parallel Nor Perpendicular - (c)

The line and plane are neither parallel nor perpendicular if the direction vector and the normal vector are neither perpendicular nor scalar multiples. Calculate the dot product: \( 1*1 + (-1)*1 + 1*1 = 1 - 1 + 1 = 1 \). Since it's not zero, the vectors are not perpendicular, and since they are not scalar multiples, they are neither parallel nor perpendicular.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Direction Vectors
A direction vector is an essential concept in vector analysis as it gives us information about the direction in which a line is oriented. For a line represented by parametric equations, like \( x = a + mt, y = b + nt, z = c + pt \), the direction vector \( \mathbf{d} \) is determined by the coefficients of \( t \). In this context, the direction vector is \( (m, n, p) \). Think of it as a kind of "slope" in three dimensions.
  • Direction vectors indicate the direction in which a line progresses as the parameter \( t \) changes.
  • The direction vector is parallel to the line, meaning it points in the same general direction the line extends.
Knowing how to identify and use direction vectors is critical, especially when analyzing the relationship between lines and planes.
Normal Vectors
Normal vectors are vectors that are perpendicular to a surface, such as a plane. For a plane defined by the equation \( ax + by + cz = d \), the normal vector \( \mathbf{n} \) is \( (a, b, c) \).
  • The normal vector is perpendicular to every line that lies within the plane.
  • It's used to define the orientation of the plane in three-dimensional space.
If you imagine the plane as a sheet of paper, the normal vector would be a pencil sticking straight up out of the paper. Understanding normal vectors helps in determining whether lines are parallel or perpendicular to planes. The relationship between the direction vector of a line and the normal vector of a plane reveals much about their spatial interaction.
Dot Product
The dot product, also known as the scalar product, is a way to multiply two vectors to get a scalar (a single number). It's crucial in determining angles between vectors and understanding vector relationships. Given two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
  • If the dot product is zero, the vectors are perpendicular.
  • A non-zero dot product indicates the vectors are not perpendicular, and if the vectors are not scalar multiples, they are not parallel.
In vector analysis, the dot product is a key tool used to explore the relationship between direction vectors of lines and normal vectors of planes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the distance between the point and the plane. $$ (0,1,5) ; 3 x+6 y-2 z-5=0 $$

Convert from spherical to rectangular coordinates. $$ \begin{array}{ll}{\text { (a) }(5, \pi / 6, \pi / 4)} & {\text { (b) }(7,0, \pi / 2)} \\ {\text { (c) }(1, \pi, 0)} & {\text { (d) }(2,3 \pi / 2, \pi / 2)}\end{array} $$

Find an equation of the plane that satisfies the stated conditions. The plane that contains the line \(x=-2+3 t, y=4+2 t\) \(z=3-t\) and is perpendicular to the plane \(x-2 y+z=5\)

Two bugs are walking along lines in 3 -space. At time \(t\) bug 1 is at the point \((x, y, z)\) on the line $$ x=4-t, \quad y=1+2 t, \quad z=2+t $$ and at the same time \(t\) bug 2 is at the point \((x, y, z)\) on the line $$ x=t, \quad y=1+t, \quad z=1+2 t $$ Assume that distance is in centimeters and that time is in minutes. (a) Find the distance between the bugs at time \(t=0\). (b) Use a graphing utility to graph the distance between the bugs as a function of time from \(t=0\) to \(t=5\). (c) What does the graph tell you about the distance between the bugs? (d) How close do the bugs get?

Find parametric equations of the line of intersection of the planes. $$ \begin{array}{l}{3 x-5 y+2 z=0} \\ {z=0}\end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.