# Collinear vectors, simulation

When I draw two collinear vectors, say $$a(2,4,6)$$ and $$b(4,8,12)$$, using a simulator, they end up being the same line and starting from the origin. They are not separated in space. The usual images of collinear vectors in textbooks show them as being separated in space.

Link to the simulator.

How do I visualise or simulate two vectors which are parallel and separated in the 3-dimensional / 2-dimensional space and not starting from the origin always?

An equivalent statement in the co-ordinate system would be, say, $$y=2x$$ and $$y=2x + 2$$ which are parallel lines. They don’t have to pass through the origin and are separated in the plane.

If vector V is denoted by $$(x,y,z)$$ then $$V=x \bar(i) + y \bar(j) + z \bar(k)$$ Where $$\bar(i), \bar(j), \bar(k)$$ are unit vectors along Cartesian axes $$X,Y,Z$$ Also. $$V=(x,y,z)$$ is actually a vector of magnitude $$|V|=\sqrt{x^2+y^2+z^2}$$and along the direction of a unit vector $$\bar(v)$$ whose Cartesian representation is $$\bar(v) = ( x/|V|,y/|V|,z/|V|)$$ Let’s look at your two vectors $$a=(2,4,6), b=(4,8,12)$$ They differ only in magnitude $$2|a|=|b|$$but have identical unit vector, $$\bar(a) , \bar(b)$$and therefore they are collinear. To plot vector a, you draw a line from origin along the direction of unit vector $$\bar(a)$$ of length $$|a|$$& similarly to plot vector $$b$$ you draw a line from origin along the direction of unit vector $$\bar(a)==\bar(b)$$ whose length is $$|b| = 2 |a|$$ they differ only in magnitude, their end tips are different,their unit vector is same. For this reason they appear along same line on simulator plot.