Determining wether a 3D line is a subspace of $R^3$

Question

Given the following two lines in 3D space:

$$ l: p_0 + t(x,y,z) \\ l_1 : (3,3,3) + t_1(1,1,1) \\ l_2: (1,1,1) + t_2(1,-1,0) $$

According to the definition, a subspace of e.g.R3 must fulfil the following criterials:

1) Passing through the origin

2) Closed under addition

3) Closed under scalar multiplication

According to the reference books, l_1 is a subspace of R3 while l_2 isn't. By drawing both lines, the first line does passing through the origin while the second doesn't. So, without drawing them out, how to determine if the 3D line is the subspace of R3? I guess that I shouldn't just determine by looking if p_0 = (0,0,0).

Answer 1

The line $l_2$ doesn't contain $(0,0,0)$ (you don't need to draw it to prove that) and therefore it is not a subspace of $\Bbb R^3$.

The line $l_1$ consists of the multiples of $(1,1,1)$ (because $(3,3,3)+t(1,1,1)=(t+3)(1,1,1)$) and so it is a subspace of $\Bbb R^3$. On any vector space $v$, if $v\in V$, the set $\{tv\mid t\in\Bbb R\}$ is a subspace of $V$.

Answer 2

If you have a line in the form $(a,b,c)+t(u,v,w)$ it is easy to determine if it passes through the origin. At least one of $u,v,w$ must be non-zero. wlog suppose $u\ne0$. Then we require $t=-a/u$ to make the first coordinate 0. Now check if it also makes the other coordinates 0.