Undergrad Plotting and visualizing a 3D plot of a vector function

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To plot the vector-valued function r(u,v) = <u + v, 3 - v, 1 + 4u + 5v>, users suggest starting by assigning the same values to u and v to simplify the process. Creating a loop for u and v can help generate corresponding x, y, and z values for plotting. Although the user acknowledges the surface is a plane, they struggle to visualize it in 3D space. The discussion highlights that any plane can be defined by three points, which can be plotted to visualize the surface. Overall, the focus is on methods to effectively plot and understand the vector function in three-dimensional space.
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Hello, I'm trying to figure out how to plot a certain vector valued function but I'm having a hard time.

The problem gives me the following vector valued function:

r(u,v) = <u + v, 3 - v, 1 + 4u + 5v>

I don't know how to plot this. So far I've tried making a table with some u and v values to get the x, y and z values so I can plot it, but it got too confusing.

Thanks in advance
 
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Where did you try it? I think you can do it by starting to giving the same value to u and v. You can try to create a loop for u and v then for each value there will be different (r,u).
 
erbilsilik said:
Where did you try it? I think you can do it by starting to giving the same value to u and v. You can try to create a loop for u and v then for each value there will be different (r,u).
Trying on paper.

sRsftzG.png


Got this table. The values of Z get too high after increasing u and v by a bit. I know the surface will be a plane, but I can't visualize it in a 3D space.
 
You know it will be a plane. Any plane is determined by three points. So just take three points, plot them and the plane through it is the one you're looking for.
 
micromass said:
You know it will be a plane. Any plane is determined by three points. So just take three points, plot them and the plane through it is the one you're looking for.
So, like this:
PFxYWGJ.jpg


?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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