Planarity of a 3D curve defined parametrically

Question

If you have a 3D curve defined parametrically by $t$ for example:

$$P=\cos\left(t+\frac{\pi}{4}\right) \frac{3}{2} {\bf i} + 3 \sin\left(t+\frac{\pi}{4}\right) {\bf j} + \cos\left(t+\frac{\pi}{4}\right) \frac{3\sqrt3}{2} {\bf k}$$

How do you determine if it lies in a certain plane?

Answer 1

A general answer is that a necessary and sufficient condition for a curve to be planar is that its torsion, given by the following formula (https://en.wikipedia.org/wiki/Torsion_of_a_curve)

$$\tau=\dfrac{\det(P'(t),P''(t),P''(t))}{\|P'(t) \times P'''(t)\|}$$

is zero. (under the condition that the third derivative exists...), itself reducible to the simpler criterion:

$$\tag{1} \det(P'(t),P''(t),P'''(t))=0.$$

For the example you give, we easily check that condition (1) is fulfilled. In fact, setting $T=t-\pi/4$ and factorizing the coefficients, we obtain:

$$\left|\begin{array}{rrr}-\sin(T)&-\cos(T)&\sin(T)\\ \cos(T)&-\sin(T)&-\cos(T)\\ -\sin(T)&-\cos(T)&\sin(T)\end{array}\right|$$

which is $0$ because rows 1 and 3 are identical.

But in this case, there is a simpler way.

It suffices to obseerve the following implication

$$\begin{cases} x= \frac{3}{2}\cos(t+\frac{\pi}{4})\\ y=3 \sin(t+\frac{\pi}{4})\\ z= \frac{3\sqrt3}{2}\cos(t+\frac{\pi}{4}) \end{cases} \ \ \ \implies \ \ \ z=\sqrt{3} \ y$$

which is translated into set langage by:

"The curve is included into the plane with equation $z=\sqrt{3} \ x.$"

Moreover, this curve is described by a periodical parametrization (see figure below).

Answer 2

If you have the plane normal direction $\bf n$ and a point on the plane $\bf r$ then the point $\bf p$ lies on the plane if

$$ {\bf n} \cdot ( {\bf p}-{\bf r} ) =0$$