Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A plane parallel to x+3y4z=7

(b) A line orthogonal to the planex+3y4z=7

(c) A plane orthogonal to the line x=3t5,y=2t+7,z=4

Short Answer

Expert verified

Part (a) The plane parallel to the plane x+3y-4z=7is 2x+6y-8z=6

Part (b) The equation of the line that is orthogonal to the plane x+3y-4z=7is r(t)=3-t,1-t,-t

Part (c) The equation of the plane that is orthogonal to the line r(t)=3t-5,-2t+7,-4is 2x+3y+5z=6

Step by step solution

01

Part (a) Step 1: Given information

The plane x+3y-4z=7

02

Part (a) Step 2: Explanation

The objective is to find the plane which is parallel to the planex+3y-4z=7

The normal vector of the plane x+3y-4z=7is N=1,3,-4

Consider the plane 2x+6y-8z=6

The normal vector of the plane 2x+6y-8z=6is N1=2,6,-8

The normal vector N1=2,6,-8can be written as:

N1=2N

Therefore, the normal vector of the plane 2x+6y-8z=6is the scalar multiple of normal vector of the plane x+3y-4z=7

Therefore, the plane parallel to the plane x+3y-4z=7is 2x+6y-8z=6

03

Part (b) Step 1: Given information

The plane x+3y-4z=7

04

Part (b) Step 2: Calculation

The objective is to find the line which is orthogonal to the plane x+3y-4z=7The normal vector of the plane x+3y-4z=7is N=1,3,-4

Consider the line r(t)=3-t,1-t,-t

The direction vector of the line r(t)=3-t,1-t,-tis d=-1,-1,-1

The dot product of d=-1,-1,-1and N=1,3,-4is:

d·N=-1,-1,-1·1,3,-4=-1-3+4=0

The dot product of d=-1,-1,-1 and N=1,3,-4 is zero. Therefore, the line and the plane are orthogonal.

Therefore, the equation of the line that is orthogonal to the plane x+3y-4z=7 is r(t)=3-t,1-t,-t

05

Part (c) Step 1: Given information

The line r(t)=3t-5,-2t+7,-4

06

Part (c) Step 2: Calculation

The objective is to find the plane orthogonal to the line r(t)=3t-5,-2t+7,-4

The direction vector of the line r(t)=3t-5,-2t+7,-4is d=3,-2,0

Consider the plane 2x+3y+5z=6

The normal vector of the plane 2x+3y+5z=6 is N=2,3,5

The dot product of d=3,-2,0 and N=2,3,5 is:

d·N=3,-2,0·2,3,5=6-6+0=0

The dot product of d=3,-2,0 and N=2,3,5 is zero. Therefore, the line and the plane are orthogonal.

Therefore, the equation of the plane that is orthogonal to the line r(t)=3t-5,-2t+7,-4 is 2x+3y+5z=6

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 Vaia!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free