Take the given vector by itself,
. Scaling the vector by a scalar
will stretch the vector. Any real number can take the place of
; when we consider all possible choices, we see that
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,
. (Upper right) By adding
to
, we get a new vector that, as we change the value of
, traces out the line through point P which also happens to be parallel to
. (Lower right)
So the equation of this line in parametric form is

or in ijk notation,
