Perpendicular Line Equations for Skew Lines in Different Planes

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SUMMARY

The discussion focuses on finding the equations for a line that is perpendicular to two skew lines represented by the equations (x/3)=(y/2)=(z/2) and (x/5)=(y/3)=(z-4)/2. Participants clarify that the cross product should be taken from the direction vectors of the lines, not the lines themselves. The correct direction vector is derived as (-2i + 4j - k), and the final equation for the perpendicular line is given as .5x - 52/7 = -.25y + 52/21 = z - 208/21, which requires adjustments to ensure intersection with the original lines.

PREREQUISITES
  • Understanding of vector calculus, specifically direction vectors.
  • Familiarity with the concept of skew lines in three-dimensional space.
  • Knowledge of the cross product operation in vector mathematics.
  • Ability to manipulate parametric equations of lines.
NEXT STEPS
  • Study the properties of skew lines in three-dimensional geometry.
  • Learn how to compute direction vectors from line equations.
  • Explore the application of the cross product in finding perpendicular vectors.
  • Review methods for deriving equations of lines from vector representations.
USEFUL FOR

Students in advanced mathematics courses, particularly those studying vector calculus and three-dimensional geometry, as well as educators looking for examples of skew line relationships and perpendicularity in space.

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Homework Statement


Find equations for a line perpendicular to both of these lines.

Homework Equations


(x/3)=(y/2)=(z/2) and (x/5)=(y/3)=(z-4)/2

The Attempt at a Solution



i don't know how to start?I know the two lines are skew, if i take the cross product will it be perp. to both lines??Can i take the cross product of two lines that lie in different planes?
 
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You don't take the cross product of the lines. You take the cross product of the direction vectors of the lines. The result is a direction vector for the perpendicular line.
 
how do i find the direction vectors?
 
Do you have a textbook? Isn't it covered in there?
 
No!My book leaves a lot of details out, it expects us to know certain calculus stuff since its a post calculus course.I just don't remember direction vectors but have studied them in the past.Ill see what i can find n google.thnx anyway.
 
Then you may need another book to keep on hand. ax=by=cz has direction vector (1/a,1/b,1/c).
 
Hmm, i did the cross product of the direction vectors and got -2i+4J-k, but the questions asks to find the equations, i got a vector.The answer in the book is .5x-52/7=-.25y+52/21=z-208/21, they surely used another method.
 
No. Look at the direction vector of the line they give as a solution. It's (1/.5,1/(-.25),1/1) which is (2,-4,1). You got the direction vector right. Now it looks like they want you to fix the constants by requiring that the perpendicular intersect the other two lines.
 

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