- #1
b0it0i
- 36
- 0
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
Last edited: