Given a curve, find the equation of the plane the cuve lies in

[b0it0i]

Homework Statement

r(t) = ( (1+t)/(1-t), 1/(1-t^2), (1/1+t) )
i have proven that the curve is planer

now the second part of the problem is to find the equation of the plane that the curve lies in

Homework Equations

Equation of a plane: A(x-xo) + B(y-yo) + C(z-zo) = 0

The Attempt at a Solution

I'm not certain how to go at this problem

Idea: I'm thinking you can choose any 3 points, t = 0, t = 2, t = 3

Then:

r(0) = (1,1,1) = A
r(2) = (-3, - 1/8, 1/4) = B
r(-2) = (-1/3, -1/3, -1) = C

Then

Vector BA = (-4, -9/8, -3/4)
Vector CA = (-4/3, -4/3/, -2)

then the normal for the plane would be BA x CA

BA x CA = (5/4, -7, 23/6)

so the equation of the plane would be

(5/4)(x-1) + (-7)(y-1) + (23/6)(z-1) = 0

thus

5/4 x - 7y + 23/6 z = -23/12or

15x - 84y +46z = -23

Idea 2: I'm thinking, since the curve is planer, the tangent at any point of the curve lies in the tangent plane

can i choose any point let's say t=4

and since the binormal is normal to the tangent plane

would I need to find B(4)

which is

T(4) x N(4)

after I have B(4), could i plug in my "normal" to the tangent plane into the equation

A(x-xo) + B(y-yo) + C(z-zo) = 0this method seems very tedious, trying to find T(t) and N(t)
Is this the correct idea(s)?
or am i totally off

If i am off, are there any hints/ suggestions? thank you

 

[b0it0i]

no takers?

 

[StatusX]

Your ideas seem fine. Why not see if they give the same answer?

 

[Guinessrulz]

I'm curious how you figured it was planer without finding T and N all ready? or for that matter the Binormal as well (using Frenets Formulas). If its planer then your curvature is a constant and your torsion is 0. That being the case you should be able to use your Binormal to find the osculating plane at your choosen point r(t). Ask if you want to hear more or you haven't already completed the problem.

 

[b0it0i]

i completed this awhile ago

i proved it was planer by using the theorem

a curve is planar if and only if the torsion is 0

i did the tedious task of computing the torsion, using the invariant formula, and it was indeed 0

thus i concluded it was planar

the easy solution was to just "guess" the plane in which the given curve lies in

assume the curve is planar
Thus the curve is spanned by gamma dot and gamma double dot
chose a point t=0

gamma dot (0)
gamma double dot (0)

took the cross product of the two, which would be the normal of the assumed tangent plane

then plug in the normal into the equation

A(x-xo) + B(y-yo) + C(z-zo) = 0

once i got that equation, i plugged in the components of gamma

x = __
y = __
z = __

into the equation of the plane, and indeed it's 0. thus gamma lies in the plane, and with this process, you also found the equation of the plane it lies in