How to determine whether a curve lies on a plane? 
Given that a sphere $x^2+y^2+z^2=1$ and a cylinder $x^2+y^2=x$ intersect at point $(1/2,1/2,1/\sqrt 2)$ determine whether the curve from the intersection lies on a plane.

We can find the curve using the following parametrization:
$$
c(t)=\bigg( \frac{1}{2}+\frac{1}{2}\cos t,\frac{1}{2}\sin t,\sin\frac{t}{2}\bigg)
$$
One of the ways to solve this is to make sure that tangent vectors to the curve are orthogonal to the plane $ax+by+cz=d$.
The tangent vector is:
$$
c'(t)=\bigg( -\frac{1}{2}\sin t,\frac{1}{2}\cos t,\frac{1}{2}\cos\frac{t}{2}\bigg)
$$
For $t=\pi$ we have $c'(t)=( 0,1/2,0)$.
Then:
$$
(a,b,c).(0,1/2,0)=0\implies b=0
$$
Similarly for $t=0\implies c=0$ and for $t=\pi/2\implies a=0$. Therefore such a plane doesn't exist.
I don't understand the logic of finding random points and why the fact that we found $a,b,c=0$ proves that the plane doesn't exist.
 A: A plane curve with non-vanishing curvature has zero torsion at all points.
The torsion $\tau$ of this curve 
$c(t)=\bigg( \frac{1}{2}+\frac{1}{2}\cos t,\frac{1}{2}\sin t,\sin\frac{t}{2}\bigg)$
is 
$\tau=\dfrac{\left({c'\times c''}\right)\cdot c'''}{\left\|{c'\times c''}\right\|^{2}}=\dfrac{12 \cos \left(\frac{t}{2}\right)}{3 \cos t+13}$
As the torsion is not identically zero, the curve is not a plane curve
Hope this helps
A: If the curve was contained on a plane $p$, let $(a,b,c)$ be a non-zero vector orthogonal to $p$. Then all vectors $c'(t)$ would be orthogonal to $(a,b,c)$ too. But the method that you described shows that there is not vector $(a,b,c)$ such that$$(\forall t\in\mathbb{R}):c'(t).(a,b,c)=0.\tag{1}$$Therefore, the curve is not contained in a plane.
There is nothing wrong with choosing random points to prove that $(1)$ is not true. The method that you described consists in supposing that such a vector exits and by, by carefully choosing certain values of $t$, to reach a contradiction. It's as if I asked: is the graph of $x\mapsto x^2$ a straight line? No, because it contains $(0,0)$, $(1,1)$ and $(2,4)$ and no straight line contains these points.
A: A plane in $\mathbb{R}^3$ has an equation of the form $a x + b y + c z = d$ where $(a, b, c) \neq (0,0,0)$. The problem of finding a plane that contains the curve can be seen as the problem of finding four such numbers $a, b, c, d$ such that
$$a x(t) + b y(t) + c z(t) = d$$
for all possible values of the parameter $t$. Usually, there are infinitely many possible values of the parameter (such as a whole interval) and each value gives a new equation that $(a, b, c, d)$ must satisfy. So we have 4 unknowns and an infinite number of equations. In most cases, there is no solution, which means that most curves don't belong to a plane.
The idea of choosing a few random points on the curve is that it may give a few independent equations that can be used to find $(a, b, c, d)$ if they exist. In most cases, 3 values of $t$ suffice to determine the plane.
Another idea is that the above equation shows that the function
$$t \mapsto a x(t) + b y(t) + c z(t)$$
must be a constant when $t$ varies. A way to check that a function is constant on an interval is to look if its derivative is $0$, so the equations imply
$$a x'(t) + b y'(t) + c z'(t) = 0$$
if the derivatives exist. Again choosing a few values $t_1, t_2, t_3$ may give equations that help find $a, b, c$ or prove that the plane does not exist.
A: First, of all, draw some pictures and you'll see that a sphere-cylinder intersection is only planar (it's a circle) when the centerline of the cylinder passes through the center of the sphere. That's not the case, here.
If you want "proof" as opposed to pictures and intuition, just use your parametric equations to compute four points $(x_i,y_i,z_i)$ on the curve. Then compute the following determinant
$$
D = \left|\begin{matrix}
x_1  &  y_1  & z_1 &  1 \\
x_2  &  y_2  & z_2 &  1 \\
x_3  &  y_3  & z_3 &  1 \\
x_4  &  y_4  & z_4 &  1 \\
\end{matrix}\right|
$$
Actually, $\tfrac16D$ is the volume of the tetrahedron having the four points as vertices, so $D$ will be zero if and only if the four points are coplanar. If you pick four points at random, you will almost certainly get a non-zero value for $D$, which will show that the curve is non-planar.
