(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 = -23/12

or

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) = 0

this 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

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

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