Question

Osculating Plane

Answer 1

The question is attached below. Anyone has an idea how to do this? i was thinking i find the T (tangent) and B (binormal) and theirs cross product will be 1 but when i put the t=0 i have nothing to solve for.

Answer 2

do as they say in the question

cross the tangent with the normal.

that vector will be the normal of the osculating plane.....and it will also be the bi-normal

Answer 3

According to
http://tutorial.math.lamar.edu/classes/calcIII/TangentNormalVectors.aspx
the tangent is the derivative of the position vector
\[ \vec{r}= <\cos t , \sin t , t> \\
\vec{T}= \vec{r'}= <-\sin t , \cos t , 1> \]
the magnitude of T is \( \sqrt{\sin^2 t+\cos^2 t + 1}= \sqrt{2}\), i.e. constant
so the normal vector N= T' will be perpendicular to T. It is
\[ \vec{N}= \vec{T'} = <-\cos t, -\sin t, 0>\]
at t=0, the two vectors are:
\[ \vec{T}= <0, 1,1>\\\vec{N}=<-1,0,0>\]
The cross-product will be the normal to the osculating plane (and the binormal)
The cross product T x N is <0,-1,1>

They say the plane contains the point at t=0, i.e.
\[ \vec{r}= <\cos t , \sin t , t> \\ \vec{r(0)}= <1,0,0>\]
Plugging into the generic formula, i.e. the plane's normal and a point:
\[ \vec{N}\cdot P= c\\
<0,-1,1>\cdot<1,0,0>= 0\]
we find the constant c is 0, and the equation of the plane is
\[ <0,-1,1>\cdot P=0\]

we can now test the choices. It looks like both A and B work.