I found a condition in a book regarding two vectors to have acute angle between them. I said that say a and b are two vectors then if they have an acute angle between them then $|a+b|>|a-b|$.

I am confused about this.. It would be great help if I get a convincing reason for this.

  • $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ Sep 19 '21 at 17:30
  • $\begingroup$ you just need to use the vector addition rule along with knowing how cosine behaves $\endgroup$ Sep 19 '21 at 17:33
  • $\begingroup$ If the vectors are in $\mathbb{R}^{n}$, and you pick a basis so that $a=|a|\vec{e}_1$, then the component of $b$ along $\vec{e}_1$ is positive if the angle between $a$ and $b$ is acute, zero if they are perpendicular, and negative if the angle is obtuse. If the component of $b$ along $\vec{e}_1$ is $b_1$, then the component of $a+b$ along $\vec{e}_1$ is $|a|+b_1$, and the component of $a-b$ along $\vec{e}_1$ is $|a|-b_1$ $\endgroup$
    – Joe
    Sep 19 '21 at 17:36

$|\vec{a}|,|\vec{b}|,|\vec{a}+\vec{b}|,|\vec{a}-\vec{b}|$ represent the sides and diagonals of a parallelogram with one vertex at origin.
In a rectangle are the diagonals congruent.
If it is not a rectangle, the figure can help. enter image description here


In case the angle between the vector $\vec a$ and $\vec b$ is acute $\cos \theta>0$, where $\theta$ is the angle between vectors $\vec a $ and $\vec b$. So, we have $\vec a .\vec b=|\vec a| |\vec b|\cos \theta >0 $.

$|\vec a+\vec b|^2 =|\vec a|^2 +|\vec b|^2 +2 (\vec a.\vec b) $

$|\vec a-\vec b|^2 =|\vec a|^2 +|\vec b|^2 -2 (\vec a.\vec b) $

Since $\vec a.\vec b>0$, we have $|\vec a+\vec b|^2 >|\vec a-\vec b|^2$ implying $|\vec a+\vec b|>|\vec a-\vec b|$

  • $\begingroup$ well, that makes sense.. Thanks I got it! :) $\endgroup$
    – Excelsior
    Sep 20 '21 at 14:42
  • $\begingroup$ Try considering accepting my answer if it is the desired one. $\endgroup$ Sep 20 '21 at 17:00

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