Intersection of two parameterized curves Disclaimer: this is homework.  I don't really understand parametrized curves in general and everything I find online in regard to this problem is in terms of $x$ and $y$, which is really throwing me off.  I need help with this problem but I'd also like an explanation of the broader picture.    
The problem asks to show that the curves parametrized by
$$r_1(t) = t\mathbf i + 2t \mathbf j + t^2\mathbf k$$
and
$$r_2(t) = t^2\mathbf i + (1-t)\mathbf j + \left(2-t^2\right)\mathbf k$$
intersect at the point $(1,2,1)$ and that the vectors tangent to the two curves at $(1,2,1)$ are perpendicular.  
 A: Cleanup first, as suggested in the comments. Then we have:
$$
\mathbf r_1(t) = t\mathbf i + 2t \mathbf j + t^2\mathbf k
= x \mathbf i + y \mathbf j + z \mathbf k \\
\mathbf r_2(u) = u^2\mathbf i + (1-u)\mathbf j + \left(2-u^2\right)\mathbf k
= x \mathbf i + y \mathbf j + z \mathbf k
$$
The intersection point is $(x,y,z) = (1,2,1)$ . Substitute:
$$
t\mathbf i + 2t \mathbf j + t^2\mathbf k
= u^2\mathbf i + (1-u)\mathbf j + \left(2-u^2\right)\mathbf k
= 1 \mathbf i + 2 \mathbf j + 1 \mathbf k \\
$$
It follows that:
$$
t = 1 \quad ; \quad 2t = 2 \quad ; \quad t^2 = 1 \\
u^2 = 1 \quad ; \quad (1-u) = 2 \quad ; \quad (2-u^2) = 1
$$
Giving as the solutions: $\,t=1\,$ and $\,u=-1$ ;
this answers the first part of the question.
For the second part, differentiate:
$$
\mathbf r_1'(t) = \mathbf i + 2\mathbf j + 2t\mathbf k \\
\mathbf r_2'(u) = 2u\mathbf i - 1\mathbf j - 2u\mathbf k
$$
And substitute the solutions for $t$ and $u$:
$$
\mathbf r_1'(1) = \mathbf i + 2\mathbf j + 2\mathbf k \\
\mathbf r_2'(-1) = -2\mathbf i - \mathbf j + 2\mathbf k
$$
The inner product of $\mathbf r_1'(1)$ and $\mathbf r_2'(-1)$ is:
$$
(\mathbf i + 2\mathbf j + 2\mathbf k) \cdot (-2\mathbf i - \mathbf j + 2\mathbf k) =
-2 - 2 + 4 = 0
$$
Thus proving that the vectors tangent to the two curves at $(1,2,1)$ are perpendicular.
