Let $r: \mathbb{R}->\mathbb{R}^3$ be a curve in arc-length parametrization such that $T(t)$ does not equal to zero for all values of $t$. Assume that the tangent line at any given point $r(t)$ of the curve always passes through a point $D$. Show that $r$ represents a straight line.
Here is my thought process on approaching this question:
For a line to be straight, the curvature $k(t$) has to be $0$.
Since it has been established that the tangent line of the curve at any given point passes through point $D$ and that $T(t)$ does not equal to $0$, the curve has to be linear.
Since the curve is linear, the tangent line, which is the derivative of the curve, has to be constant.
Because I have to show that the tangent line is straight, I have to show that the curvature of this tangent line is $0$. The tangent line $T(t)$ for $r(t)$ is zero and the derivative of $T(t)$ is also zero, therefore the curvature of $r(t)$ is zero.
Can someone please tell me if I am right?
