Simple question of vectors and points

Question

Let $L_0$ be a line in $R^3$, that passes through the points $(1, 2, 3)$ $(0, 0, 0 )$

Let $L_1$ be the parallel line to $L_0$ that passes through $(3, 2, 1)$

Prove/disprove:

$(1,1,1) \in L_1$

$(-1,-2,-3) \in L_1$

I got this problem in my linear algebra course and I'm not quite sure what is the subject of this material, I tried to dig through vector spaces but couldn't tell how to solve this problem.

Answer 1

Hint:

Every line can be represented as $$ L=p_{0}+tp_{1} $$

with $t\in\mathbb{R}$.

In the case of $L_{0}$ we have $p_{0}=0$, $p_{1}=(1,2,3)$ $$ L_{0}=t(1,2,3) $$

$$ L_{1}=(3,2,1)+t(1,2,3) $$

since it is parallel to $L_{0}$ and passes through $(3,2,1)$.

Answer 2

A parametrization of $L_0$ is $(t,2t,3t)$, so that $t=0$ gives $(0,0,0)$ and $t=1$ gives $(1,2,3)$.

Then a parametrization of $L_1$ is $(t+3, 2t+2,3t+1)$. There is no $t$ that gives $(1,1,1)$ or $(-1,-2,-3)$ for $L_1$. Indeed, if $t+3=1$, then $t=-2$, but then $2t+2=-2\ne1$. Similarly, if $t+3=-1$, then $t=-4$, then $2t+2=-6\ne-2$.

Therefore, none of the two points belongs to $L_1$.