# prove the angle of sum of vectors is always in the middle

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.

• What have you tried? Aug 30, 2020 at 18:11
• 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. Aug 30, 2020 at 18:26
• 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). Aug 30, 2020 at 18:48
• I think I’m convinced, and I was looking for something concise and rigours Aug 30, 2020 at 18:50
• $v_3$ 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.
– Moti
Aug 30, 2020 at 19:25

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 $$a_1/b_1 (because $$\angle v_1<\angle v_2$$), can you prove the above inequality?