Take the given vector by itself, [tex]\mathbf v=2\,\mathbf i+\mathbf j-\mathbf k[/tex]. Scaling the vector by a scalar [tex]t[/tex] will stretch the vector. Any real number can take the place of [tex]t[/tex]; when we consider all possible choices, we see that [tex]t\mathbf v[/tex] traces out an infinite line through the origin and the point (2, 1, -1). (Left)
Next, treat point P(2, 1, -3) as a vector, [tex]\mathbf P=2\,\mathbf i+\mathbf j-3\,\mathbf k[/tex]. (Upper right) By adding [tex]\mathbf P[/tex] to [tex]t\mathbf v[/tex], we get a new vector that, as we change the value of [tex]t[/tex], traces out the line through point P which also happens to be parallel to [tex]\mathbf v[/tex]. (Lower right)
So the equation of this line in parametric form is
[tex]\mathbf r(t)=(2,1,-3)+t(2,1,-1)=(2+2t,1+t,-3-t)[/tex]
or in ijk notation,
[tex]\mathbf r(t)=(2+2t)\,\mathbf i+(1+t)\,\mathbf j-(3+t)\,\mathbf k[/tex]
