 Solution

The line through the origin is $l_0=\mathrm{span}(u)$, containing zero and closed under addition and scalar multiplication, so it is a vector subspace. If $l_{x_1}=l_{x_2}$, then $x_1\in l_{x_2}$ and $x_1=x_2+\lambda u$ for some real $\lambda$. Conversely that equality gives the same set after shifting the line parameter. ThusChanging representatives to $x^{\prime }=x+su$, $y^{\prime }=y+tu$ changes their sum by $(s+t)u$ and a scalar multiple by $\alpha su$. The line classes therefore do not change, proving both operations well-defined. Associativity follows from $l_{(x+y)+z}=l_{x+(y+z)}$. The zero vector of this quotient vector space is the entire line $l_0$, because $l_x+l_0=l_x$.

For a basis $(u,b_1,b_2)$ of ${\mathbb{R}}^3$, every vector is $x=su+t{b}_1+v{b}_2$, so $l_x=t{l}_{b_1}+v{l}_{b_2}$. If this combination equals $l_0$, then $t{b}_1+v{b}_2=su$; independence of the three original basis vectors forces $t=v=s=0$. Hence $(l_{b_1},l_{b_2})$ is a basis of the space of parallel lines as a quotient vector space.

For the specified direction, add $zu$ to $(x,y,z{)}^T$ to obtain the representative $(x+z,y+3z,0{)}^T$. Thus its coordinates in the two standard line classes are $(x+z,y+3z{)}^T$. The two supplied image classes have coordinates $(1,1{)}^T$ and $(-3,1{)}^T$, while their input coordinates are $(1,-1{)}^T$ and $(1,1{)}^T$. If $a_1,a_2$ are the matrix columns, then $a_1-a_2=(1,1{)}^T$ and $a_1+a_2=(-3,1{)}^T$. Solving gives