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For two planar vectors $v_1, v_2$, and they are in the same quadrant.

Define $\angle(v_1)$ as the angle between $v_1$ and positive x-axis. And $\angle{v_1} \lt \angle(v_2)$.

Their sum $v_3 = v_1+v_2$, would this be true $\angle(v_1) \lt \angle(v_3) \lt \angle(v_2)$? can you prove it?

Here is what I have tried:

It should be easy to see $v_3$ is in the same quadrant with $v_1, v_2$, since $v_3= (a_3, b_3)=(a_1+a_2, b_1+b_2)$, which $\forall a_i$ have the same sign and $\forall b_i$ have the same sign.

Since all $v_1, v_2, v_3$ are in the same quadrant, the angle between any two of them is less than 90. The three of them would form an acute triangle.

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  • $\begingroup$ What have you tried? $\endgroup$ Aug 30, 2020 at 18:11
  • $\begingroup$ thanks for your interest, i did try to draw them on paper, and nothing concrete comes to me :( i think i only have the first step $v_3$ is also going to be in the same quadrant. $\endgroup$
    – peng yu
    Aug 30, 2020 at 18:26
  • $\begingroup$ If the drawing of $v_1 + v_2$ and $v_2+v_1$ (both being equal of course) doesn't convince you and you want something more "coordinaty", try comparing the angle tangents $a_1/b_1$, $a_2/b_2$ and $a_3/b_3$ (without loss of generality you can consider all your vectors to be in the upper right quadrant). $\endgroup$
    – Tassle
    Aug 30, 2020 at 18:48
  • $\begingroup$ I think I’m convinced, and I was looking for something concise and rigours $\endgroup$
    – peng yu
    Aug 30, 2020 at 18:50
  • $\begingroup$ $v_3$ is along a diagonal of a parallelogram. You could rotate the parallelogram and based on your definition see that it might get values that are no fitting you claim - unless the direction of the vectors have some limitations. $\endgroup$
    – Moti
    Aug 30, 2020 at 19:25

1 Answer 1

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Hint (elaborating on Tassle's comment):

You have that $$\tan\angle v_1=\frac{a_1}{b_1},\ \tan\angle v_2=\frac{a_2}{b_2},\ \tan\angle v_3=\frac{a_1+a_2}{b_1+b_2}.$$ Since the $\tan$ function is monotonically increasing, it suffices to show that $$\frac{a_1}{b_1}<\frac{a_1+a_2}{b_1+b_2}<\frac{a_2}{b_2}.$$ Since we know that $a_1/b_1<a_2/b_2$ (because $\angle v_1<\angle v_2$), can you prove the above inequality?

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