Parametric Surfaces and Their Areas

[wubie]

[SOLVED] Parametric Surfaces and Their Areas

Hello,

I am having problems visualizing a concept. First I will post my question as it is given in Jame's Stewart's Fourth Edition Multivariable Calculus text, Chapter 17, section 6, question 17.

Find a parametric representation for the given surface.

(a) The plane that passes through the point (1,2,-3) and contains the two vectors i + j + - k and i - j + k .

Now I know that vector representation in the solution can be written as

r(u,v) = rsub0 + ua + vb

where a = i + j + - k and b = i - j + k which becomes

r(u,v) = <1,2,-3> + u<1,1,-1> + v<1,-1,1>

which would produce parametric equations

x = 1 + u + v,
y = 2 + u -v,
z = -3 -u + v.

But what I am wondering what if I let a = i - j + k and b = i + j + - k . Then I would have

r(u,v) = <1,2,-3> + u<1,-1,1> + v<1,1,-1>

which would produce different parametric equations than the first.

x = 1 + u + v,
y = 2 - u + v,
z = -3 + u - v.

Now intuitively I think this is just as valid as the first. Is it though?

Any help / input is appreciated. Thankyou.

I'm back and the more I think about it and fool around with it I believe it is not possible to have two vector equations representing a plane with the above criteria. If that is the case then how do I determine which vector is multiplied by the parameter u and which vector is multiplied by the parameter v? This is the part that is confusing me.

 

[matt grime]

u and v are just dummy (free) variables. the two planes are the same.

Perhaps it's easier to see:

just take the x,y plane itself, in R^2

then all points on it can be described as the set {(u,v) | u,v in R} and is equally the set {(s,t) | s,t in R} and hence, {(v,u) | v,u in R}

nothing special going on.

 

[wubie]

I think I know what you are saying, but I am not following

then all points on it can be described as the set {(u,v) | u,v in R} and is equally the set {(s,t) | s,t in R} and hence, {(v,u) | v,u in R}

Particularly when you go from

is equally the set {(s,t) | s,t in R} and hence, {(v,u) | v,u in R}

Usually it is the simplest things that stump me. Over thinking too much perhaps? I don't know.

 

[Chen]

The two planes are the same just like these two lines are the same:
$$\underline{u} = (1, 2, 3) + \alpha (4, 5, 6)$$
$$\underline{v} = (5, 7, 9) + \beta (8, 10, 12)$$
When you combine two vectors linearly to form a surface, the scalar coefficients can take absolutely any value and they have no special meaning...

 

[matt grime]

To echo Chen, the u,v,s,t are just parameters free to roam over any (real) value. The labels have no intrinsic meanings.take the even numbers, then it is the set of all objects, s where s=2t for some integer t. I picked those labels at random and could have equally put them in the other order. It isn't important as long as you don't mix things...

another example:

the definite integral from 0 to 1 of f(x)dx is the same as if we did it for f(y)dy, or f(s)ds...