Csdrhrt

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.

I am just expanding memerson's comment:

Part 1

The tangent vectors at time $t_1$ and $t_2$ are parallel if:
$$
begin{pmatrix}
1 \
2t_1-1 \
-2t_1 \
end{pmatrix}
=lambda
begin{pmatrix}
1 \
2t_2-1 \
-2t_2 \
end{pmatrix}$$

as you can see this is simply impossible if $lambdaneq1$ and $lambda=1implies$ $t_1$ and $t_2$ are the same time point.

Part 2
The tangent vector at any point is given by
$$
begin{pmatrix}
1 \
2t-1 \
-2t \
end{pmatrix}$$
at $x(2)$ the tangent vector is given by $(1,3-4)$. Therefore we need to find a time $t_1$ such that the tangents are perpendicular or the inner product(dot product) is zero.
$$(1,3,-4)cdot(1,2t_1-1,-2t_1)=0\
implies 1 +6t_1 -3 +8t_1=0\
implies t_1=frac{1}{7}
$$