Vector Parallel to the Line

Your exact same problem is solved here in Example 2.

Think about the vector ⟨5, 2-t, 10+4t⟩: 5, 2 and 10 are just offsets.

Let's consider a simpler case: y = mx + b, a line expressed in slope-intercept form.

Say we've got two lines:

y = 5x + 2

and

y = 5x + 8

Are they parallel even though their y-intercepts are different? Sure they are, because their slopes are the same. Changing where they intercept the y-axis doesn't change this fact. All b does is translate the line vertically. The lines are still parallel. And since we're only interested in whether they're parallel or not, we can ignore b entirely to get y = mx.

In the three-dimensional case we have here, it is no different except that the idea of slope is little more complex, but it is still a coefficient on each component just like 'm' is for the slope-intercept line. Similarly, adding or subtracting a displacement to each component does not affect it's 'slope', only its displacement. And since only its 'slope' figures in determining whether it is parallel to another vector, we can ignore the displacements entirely. All they do is just move the vector around, they don't affect its 'slope.' Only the coefficients do, and since this vector parameterised, we can plug in any value we want for 't' and so we can fix t = 1 and voila! ⟨0,-1,4⟩! Dig?

Nobody's perfect,you cannot remember everything from highschool mate