Why $k\mathbf{v}$ points in the same direction as $(kv_1,kv_2)$

Question

I understand everything from the book excerpt below except why it points in the same direction.

How's the direction here shown? Because it is the original vector simply multiplied by (a positive) scalar?

Answer 1

$k=0$

$\large k\vec v_1=k\vec v_2=(k.0,k.0)=(0,0)$

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$k\ne0$

Slope of line will be $$\large\frac{k\vec v_2}{k\vec v_1}=\frac{\vec v_2}{\vec v_1}$$ same as of vectors.This doesn't mean that it is on same side of origin on the line of this slope or, what you say, $k>0$. It simply means if $k>0$ it'll be on same side of origin on this line and vice-versa for $k<0$.

Answer 2

Here is one way to think about it. Consider a right triangle with one leg parallel to the $x$-axis having length $v_1 > 0$ and one leg parallel to the $y$-axis having length $v_2 > 0$. The direction of the hypotenuse can be found by finding the angle $\theta$ between the hypotenuse and the leg $v_1$, which will be $\theta = \tan^{-1}(v_2/v_1)$. Now, if we scale $v_1$ to $k v_1$ and if we scale $v_2$ to $k v_2$, then the angle $\phi$ between the leg $kv_1$ and the new hypotenuse (after scaling) will be $\phi = \tan^{-1}(kv_2/(kv_1)) = \tan^{-1}(v_2/v_1) = \theta$.