Perpendicular Line Equations for Skew Lines in Different Planes

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Homework Help Overview

The discussion revolves around finding equations for a line that is perpendicular to two given skew lines represented in parametric form. The subject area includes vector calculus and geometry, particularly focusing on the properties of lines in three-dimensional space.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of skew lines and the method of using cross products to find a perpendicular line. Questions arise about the validity of taking the cross product of lines in different planes and how to derive direction vectors from the given line equations.

Discussion Status

Participants are actively engaging with the problem, with some offering guidance on finding direction vectors and others expressing uncertainty about the material. There is a recognition of the need to clarify the relationship between direction vectors and the equations of the lines.

Contextual Notes

One participant notes that their textbook lacks detail on certain topics, indicating a potential gap in foundational knowledge necessary for solving the problem. The discussion also highlights the complexity of the problem due to the post-calculus level of the course.

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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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