How to find the equations of the normal & osculating plane of a curve?

Practicing for my finals!



I have a question "Find the equations of the normal plane and osculating plane of the curve at the point (1,1,1)"



How do I go about this question? This is all that is given.



Thank you!How to find the equations of the normal %26amp; osculating plane of a curve?This can't be all, because no curve is given! So it seems that your professor is looking for a description of the method you would use to find those equations, once you're given a curve.

So suppose the curve, C, is given parametrically by 3 functions of a parameter, t:
C: (x,y,z) = (f(t), g(t), h(t))

and that for some value of t, say, t=t0, all 3 functions take the value 1:
(f(t0), g(t0), h(t0)) = (1, 1, 1),
so that the curve will pass through that point.

Then you can use the fact that the normal plane is perpendicular to the tangent vector to C, and that the osculating plane contains both the tangent vector and its derivative.

The tangent vector, T_, is any non-zero scalar multiple of
dC/dt = (f'(t), g'(t), h'(t))

while any non-zero scalar multiple of its derivative, T'_
K_ = cT'_ = c(f"(t), g"(t), h"(t)); c 鈮?0
is non-collinear with T_, and K_ and T_ together span the osculating plane.

T'_ will be the curvature vector, and perpendicular to T_, if the parameter, t, is the distance, s, along the curve, or any non-zero scalar multiple of s. Generally, for these methods to work, t must be monotonically increasing or decreasing with s; it can't 'stall', or 'reverse direction' along the curve.
***
EDIT: a little further analysis shows that this takes the mathematical form, that taking t as a function of s, then we must have dt/ds 鈮?0, everywhere along the curve.
***

A good example is the helix given by:
x = f(t) = a cos t
y = g(t) = a sin t
z = h(t) = bt
Taking a = 鈭? and b = 4/蟺, it will pass through (1, 1, 1) at t = 蟺/4. (t here is proportional to s.)

So the normal plane is one which is normal to the vector T_ and passes through the point (1, 1, 1)
The osculating plane is parallel to the plane spanned by the 2 vectors T_ and K_, and passes through the point (1, 1, 1).

To do this you'll need the following techniques of linear algebra:

The equation of a plane passing through the point P = (u, v, w), and normal to the vector
A_ = (a, b, c), calling r_ the position vector, (x, y, z), is
A_鈥_ = A_鈥, that is,
ax + by + cz = au + bv + cw

** Note that the LHS of this equation is a linear expression in the variables x, y, %26amp; z; and the RHS is a constant. **

The equation of a plane parallel to the two independent vectors
A_ = (a, b, c), and
D_ = (d, e, f),
and passing through the point P = (u, v, w)
is (both sides of the "=" are determinants):

| a b c | . | a b c |
| d e f | = | d e f |
| x y z | . | u v w |

This can be written in terms of vector triple products:
A_xD_鈥_ = A_xD_鈥

** Note that, again, the LHS is a linear expression in x, y, %26amp; z; and the RHS is a constant. **How to find the equations of the normal %26amp; osculating plane of a curve?What curve?