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

[Lancelot59]

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

e^{x}sin(y)=2 at the point (ln(2),\frac{\pi}{2}). 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?

 

[dacruick]

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.

 

[SammyS]

How is 3x-2y-4 a line?

 

[dacruick]

How is 3x-2y-4 a line?

...?

 

[SammyS]

...?

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.

 

[dacruick]

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.

 

[Lancelot59]

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.