# How to check for acute angle between two vectors?

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.

• you just need to use the vector addition rule along with knowing how cosine behaves Sep 19 '21 at 17:33
• 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 $|a|-b_1$
– 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.

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 $$|\vec a+\vec b|>|\vec a-\vec b|$$

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