Finding Normal & Tangent Vectors to Line: 3x-2y-4

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Homework Help Overview

The discussion revolves around finding normal and tangent vectors to the line represented by the expression 3x-2y-4. Participants explore the implications of the expression and its interpretation as an equation of a line.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive normal and tangent vectors but expresses uncertainty about the validity of their method. They also reference a different function and question their approach. Some participants question the nature of the expression as a line and clarify the correct form of the equation.

Discussion Status

Participants are actively questioning the assumptions regarding the expression as a line and discussing the necessary conditions for it to be considered an equation. There is a mix of interpretations being explored, particularly regarding the correct formulation of the line.

Contextual Notes

There is confusion regarding the expression 3x-2y-4 and its status as an equation, with participants noting the importance of including "=0" for it to represent a line. This highlights a potential misunderstanding in the problem setup.

Lancelot59
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I figured this would be easy, I need to find the normal and tangent vectors to this line:

3x-2y-4

Well simple enough, I got the correct parametric equations for the normal, but the tangent line is being silly. I dumbed it out and got the right answer, but I think it was for the wrong reason.

I just treated the coefficients as the slope for the other variable, and it worked. I know that won't work all the time though. As shown by the next problem I did where the function was

[tex]e^{x}sin(y)=2[/tex] at the point [tex](ln(2),\frac{\pi}{2})[/tex]. I got (2,0) for the gradient at the point, and then did the same thing.

I got the correct parametric functions, but the method seems flawed. What's the proper way to do this?
 
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hi,

you should think more about what you're doing instead of relying on formulas so heavily.

finding the tangent line to a line will be pretty trivial.

As far as the normal vector is concerned, you shouldn't have to look too far to find a solution. There are a few ways, and I suggest searching google first.
 
How is 3x-2y-4 a line?
 
SammyS said:
How is 3x-2y-4 a line?
...?
 
dacruick said:
...?
3x-2y-4 is not an equation!

3x-2y-4 = 0 is the equation of a line.

z = 3x-2y-4 is the equation of a plane.
 
SammyS said:
3x-2y-4 is not an equation!

3x-2y-4 = 0 is the equation of a line.

z = 3x-2y-4 is the equation of a plane.

Haha right you are. I honestly assumed there was an "=0". I think my brain automatically put it there.
 
dacruick said:
Haha right you are. I honestly assumed there was an "=0". I think my brain automatically put it there.

Mine did too when I made the post.
 

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