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.
Best Answer
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