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.

  • $\begingroup$ Define "angle". $\endgroup$ – Kenny Lau Oct 6 '17 at 21:01
  • $\begingroup$ (I'm saying this because the theorem comes directly from the definition of angle) $\endgroup$ – Kenny Lau Oct 6 '17 at 21:02
  • $\begingroup$ Care to clarify? The angle would be the space between the vectors at their intersecting point. $\endgroup$ – DreaDk Oct 6 '17 at 21:07
  • $\begingroup$ Define it in terms of the vectors only. $\endgroup$ – 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 $|\vec a+\vec b| > \sqrt{|\vec a|^2+|\vec b|^2}$.

| cite | improve this answer | |

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 $|\vec{a}+\vec{b}|^2>|b|^2$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.