# Sum of vectors theorem

Prove the following theorem:

If the angle between the vectors $\vec a$ and $\vec b$ is not greater than 90 degrees then the length of the vector ($\vec a + \vec b)$ is not less than $\vec a$ or $\vec b$.

The theorem is in the part where only vector addition is introduced so would appreciate if someone can post a solution involving only summing vectors. Otherwise any other solution is welcome. I tried using triangle inequality/law of cosines but they both involve vector products.

PS: Constructing random vectors it seems to be that the smaller the angle the smaller the sum of the two vectors is going to be.

• Define "angle". – Kenny Lau Oct 6 '17 at 21:01
• (I'm saying this because the theorem comes directly from the definition of angle) – Kenny Lau Oct 6 '17 at 21:02
• Care to clarify? The angle would be the space between the vectors at their intersecting point. – DreaDk Oct 6 '17 at 21:07
• Define it in terms of the vectors only. – Kenny Lau Oct 6 '17 at 21:07

The angle $\theta$ of $\vec a$ and $\vec b$ is defined as the unique real number in $[0,\pi]$ such that $\cos \theta = \dfrac{\vec a \cdot \vec b}{|\vec a| |\vec b|}$.

When $\theta<90^\circ$, we have $\vec a \cdot \vec b > 0$.

Therefore, $|\vec a+\vec b|^2 = (\vec a+\vec b)\cdot(\vec a+\vec b) = |\vec a|^2 + 2 \vec a \cdot \vec b + |\vec b|^2 > |\vec a|^2 + |\vec b|^2$.

Therefore, $|\vec a+\vec b| > |\vec a|$ and $|\vec a+\vec b| > |\vec b|$.

A stronger lower bound would be $|\vec a+\vec b| > \sqrt{|\vec a|^2+|\vec b|^2}$.

Let $\vec{a}=(a_x,a_y)$ and $\vec{b}=(b_x,b_y)$, so $\vec{a}+\vec{b}=(a_x+b_x,a_y+b_y)$, so it's length is $|\vec{a}+\vec{b}|=\sqrt{(a_x+b_x)^2+(a_y+b_y)^2}$, and $|\vec{a}+\vec{b}|^2=(a_x+b_x)^2+(a_y+b_y)^2=a_x^2+2a_xb_x+b_x^2+a_y^2+2a_yb_y+b_y^2=|a|^2+|b|^2+2(a_xb_x+a_yby)=|a|^2+|b|^2+2(\vec{a}\cdot\vec{b})$

Since the angle between both vectors is less than $90º$, then $\vec{a}\cdot\vec{b}>0$, so $|\vec{a}+\vec{b}|^2>|a|^2+|b|^2$, so $|\vec{a}+\vec{b}|^2>|a|^2$ and $|\vec{a}+\vec{b}|^2>|b|^2$