Determine if the given point lies on the parametric line.

[KnowledgeisPo]

Homework Statement

Determine whether the points P(7,10,4) and Q(5,22,5) lie on the given line:

Homework Equations

r(u,v)=<2u+3v, 1+5u-v, 2+u+v>

The Attempt at a Solution

x=x(u,v)=2u+3v, y=1+5u-v, z=2+u+v
x+y+z=8u+3v+3

 

[LCKurtz]

Homework Statement

Determine whether the points P(7,10,4) and Q(5,22,5) lie on the given line:

Homework Equations

r(u,v)=<2u+3v, 1+5u-v, 2+u+v>

The Attempt at a Solution

x=x(u,v)=2u+3v, y=1+5u-v, z=2+u+v
x+y+z=8u+3v+3

The parametric equation of a line has only one parameter:

R(t) = <x(t),y(t),z(t)>

Start by figuring out the equation. Or is there a "given line" or point you haven't mentioned? Or are you given a surface??

 

[KnowledgeisPo]

I see what you are saying. Keep in mind this is Parametric Surfaces and Their Areas. That was all the given information, and that is how the wording is transcribed in the book. Perhaps P and Q corresponds to the vector itself, and i and j of the vector are to correspond to P and Q respectively, for z to confirm proper coordinates. The exact wording is as follows:

"Determine whether the points P and Q lie on the given line.
1. r(u,v) = <2u+3v, 1+5u-v, 2+u+v>
P(7,10,4), Q(5,22,5)"

 

[KnowledgeisPo]

Solved

Thank you for your time, and sorry for any inconvenience, as I solved it myself. I set the partial vectors equal to the respective points, and used the simple Z partial to put v in terms of u. Upon doing that, I substituted that in for the X partial to get u and v coordinates at the point. Upon applying the u and v coordinates to the Y partial, the equation of the Y partial equaled to the points came out true and false for each of the points. The true and false calculations matched the answer in the back of the book.

Thank you, and if you request, I will post the solution.

 

[LCKurtz]

I see what you are saying. Keep in mind this is Parametric Surfaces and Their Areas. That was all the given information, and that is how the wording is transcribed in the book. Perhaps P and Q corresponds to the vector itself, and i and j of the vector are to correspond to P and Q respectively, for z to confirm proper coordinates. The exact wording is as follows:

"Determine whether the points P and Q lie on the given line.
1. r(u,v) = <2u+3v, 1+5u-v, 2+u+v>
P(7,10,4), Q(5,22,5)"

But that R(u,v) is a surface, not a line. It would make sense to check whether those two points are on the surface.

Thank you for your time, and sorry for any inconvenience, as I solved it myself. I set the partial vectors equal to the respective points, and used the simple Z partial to put v in terms of u. Upon doing that, I substituted that in for the X partial to get u and v coordinates at the point. Upon applying the u and v coordinates to the Y partial, the equation of the Y partial equaled to the points came out true and false for each of the points. The true and false calculations matched the answer in the back of the book.

Thank you, and if you request, I will post the solution.

And whether or not the points lie on the surface has nothing to do with the partial derivatives of R. So at this point, I still have no idea of what you are trying to do.

 

[KnowledgeisPo]

But that R(u,v) is a surface, not a line. It would make sense to check whether those two points are on the surface.

Yes, this is a little bit why I didn't understand the book.

And whether or not the points lie on the surface has nothing to do with the partial derivatives of R. So at this point, I still have no idea of what you are trying to do.

I improvised on terminology a little bit when I said partial vectors, then continued to allocate them to I,J, and K components of what would be a function on a higher integration.

 

[HallsofIvy]

"Determine whether the points P and Q lie on the given line.
1. r(u,v) = <2u+3v, 1+5u-v, 2+u+v>
P(7,10,4), Q(5,22,5)"

As has been said, that is a surface (actually a plane), not a line. To determine if the point (7, 10, 4) lies on that plane, just set 2u+ 3v= 7, 1+ 5u- v= 10 and solve for u and v. If those values of u and v also satisfy 2+ u+ v= 4, then the point is on the plane.

 

[KnowledgeisPo]

Yes. As I said a few posts, I have solved it. I was aware that multivariable parametric surfaces are surfaces. However, was I supposed to not say what the book said? Thank you again for the assistance.

 

[LCKurtz]

Yes. As I said a few posts, I have solved it. I was aware that multivariable parametric surfaces are surfaces. However, was I supposed to not say what the book said? Thank you again for the assistance.

You're welcome and we are glad you solved your problem, whatever it was. In an odd sort of way, I would like to see your solution to see if I could figure out what the problem was.

 

[KnowledgeisPo]

...and if you request, I will post the solution.

<2u+3v, 1+5u-v, 2+u+v>
P(7,10,4) Q(5,22,5)

P: 7=2u+3v, 10=1+5u-v, *4=2+u+v*
2=u+v, so u=2-v
7=2(2-v)+3v
7=4-2v+3v
3 =v,, 7=2u+9, u=-1

10=?=1+5(-1)-3,, 10=/=-7

Q: 5=2u+3v, 22=1+5u-v, *5=2+u+v*
3=u+v, so u=3-v
5=2(3-v)+3v
5=6-2v+3v
-1=v,, 5=2u-3, u=4

22=?=1+5(4)-(-1),, 22=1+20+1