Find parallel vector with unknown variable I'm having some difficulties understanding this:

In $\Bbb R^3$ equipped with the inner product $\langle.,. \rangle$,
let $\vec{AB}=(1,-3,-2), C=(−1,2,1) $
and for$t\in \mathbb R, X(t)=(0,t,t)$. Find $t^*\in \Bbb R$ such that $\overrightarrow{CX(t^*)}$ is parallel with $\vec{AB}$.

My understanding is that if a vector is to be parallel, the following is true:
$$\langle(1,-3,-2),(0-(-1),t-2,t-1)\rangle = 0$$
Here I have found out that $t^*=-1$, but if you insert $t^*=-1$, you get that $\vec{AB} = \overrightarrow{CX(t^*)}$. Are the vectors parallel? I understand that they are the same not parallel?
 A: Just a small remark:

*

*Suppose that $u,v$ are vectors in the space ${\bf R}^{n}$, then by definition $u$ and $v$ are parallels vectors if there exists a scalar $k$ non zero such that $u=kv$.


*Suppose that $u,v\in {\bf R}^{n}$ are no zero vectors, then if $u\cdot v=0$, then $u$ and $v$ they are orthogonal vectors.

There are some things that are not clear in what you wrote. You write that since they are parallel, the dot product between the vector $(1,-3,-2)$ and the vector $(1,t-2,t-1)$ is $0$, but that is not true. If you impose the dot product equals to zero, then you arrived orthogonal vectors as I said in the remark. Now, you said that "here I have found out that $t=−1$" but using the dot product equals to zero to arrived to $t=9/5$ which is the value of $t$ for what $\vec{CX}$ and $\vec{AB}$ are orthogonal vectors. On other hand if you define $t=-1$, then $(1,t-2,t-1)$ is just $(1,-3,-2)$ and then indeed $\vec{CX}=(1)\vec{AB}$ and then by definition they are parallels vector if you setting $t=-1$.
A: A major mathematical step when you do not understand something in mathematics is to simplify. As stated in the comments:

