We have $$x(t)= \begin{pmatrix} 1+t \\ t^2-t \\ 1-t^2 \\ \end{pmatrix}$$
Part 1
Are there $2$ points $x(t_1)$, $x(t_2)$, such that the function’s tangent vectors at these points are parallel to each other? Find such points, or show that none exist.
$$x'(t)= \begin{pmatrix} 1 \\ 2t-1 \\ -2t\\ \end{pmatrix}$$
I tried to search online but how can I prove that they are parallel to each other if I can't even find the points? I got stumped at $2t^2-2t+1=0$
Basically I get a negative in a square root. And that would lead to no solution for t. My answer is wrong... or is the answer scheme wrong?
Part 2
Given the point $x(2)$, is there a second point such that the tangent vectors are perpendicular to each other? Find such a point or show that it doesn’t exist.
How can I find the second point? And to prove that they are perpendicular do I have to differentiate twice?
I would appreciate some guidance.