Question

Binormal

Answer 1

I have a space curve (x,y,z) = (t, t^2, t^3) and I'm supposed to find a point through which its binormal at (1,1,1) passes.
So, I found out the binormal equation through \[B = \frac{ r' X r'' }{ \left| r' X r'' \right| }\]
and got \[B = \frac{ \left( 6t ^{2}, -6t, 2 \right) }{ \sqrt{36t ^{4}+36t ^{2} + 4}}\]
and now I'm pretty much stuck here. Any idea about what to do next?

Answer 2

P.S. r = (t, t^2, t^3)

Answer 3

i reckon the formulae you are using are not what you think they are. i've found them on Wiki but they're making assumptions.

it's also not clear what you are actually trying to achieve. because any line you like passes through (1,1,1). including the line you think you have found.

FTR, you can de-param that vector line into a closed curve as \(z(x,y) = xy\) etc and find the normal that way.

not helpful, i suspect :-(

Answer 4

@IrishBoy123 i have my binormal as i said above and the binormal is supposed to pass through 2 points, the first one being on the curve at (1,1,1) and then there's the other point that i have to find but i don't know how.

if it helps, the options in the MC were (A) (4, −2, 2), (B) (7, 1, −1), (C) (6, 0, 0), (D) (8, −1, 0), (E) (4, 1, 0).

Answer 5

The point (1,1,1) occurs at t= 1 so pop that unto your binormal vector to get \(\vec B = (3,-3,1)\),

your parameterised line through (1,1,1) is then \(\vec l(s) = (1,1,1) + s (3, -3, 1)\)

and \(\vec l(1) = (4,-2,2)\) which is option A

[your formula for \(\vec B\) seems to work though i am not totally sure why. i did it differently and got the same answer]

Answer 6

@IrishBoy123 i'm so sorry to disturb you again and again but how do you get B⃗ =(3,−3,1)? I'm getting \[B = \left( \frac{ 6 }{ \sqrt{76} }, \frac{ -6 }{ \sqrt{76}}, \frac{ 2 }{ \sqrt{76} } \right)\]

Answer 7

yes, we re both right!! the thing is, all you need from that \(\vec B\) vector is its direction, yes?

so we can either write the line as

\( \vec l = (1,1,1) + s \left( \frac{ 6 }{ \sqrt{76} }, \frac{ -6 }{ \sqrt{76}}, \frac{ 2 }{ \sqrt{76} } \right) \)

which looks like crap

......or as

\( \vec l = (1,1,1) + s (3,-3,1)\)

same thing, but the second one is just a lot easier to play with

Answer 8

oooh okay! @IrishBoy123 Thanks a lot!