Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given 2 vectors in $\mathbb{R}^3$, $A$ and $B$, and the condition $\|A-B\|=\|A\|$. How can I find the angles that the vectors A and B could form?

I've started with: $\cos\theta=\frac{AB}{\|A\|\cdot\|B\|}$, and trying to derive some trick to relate the angle between $A-B$ and $B$ but it leads me nowhere.

share|cite|improve this question
Every acute angle is possible. Draw a picture in 2D. – Did Nov 8 '12 at 7:11
up vote 3 down vote accepted

Choose an orthonormal basis $(\vec u,\vec v)$ in the $(A,B)$ plane, assume without loss of generality that $A=a\vec u$ for some $a\gt0$ and $B=b\cos\theta \vec u+b\sin\theta\vec v$ for some $|\theta|\leqslant\frac\pi2$ and $b\ne0$. Then $\|A\|=\|A-B\|$ is equivalent to $a^2=(a-b\cos\theta)^2+(b\sin\theta)^2$, that is, $b=2a\cos(\theta)$.

One sees that every angle $|\theta|\lt\frac\pi2$ yields a vector $B=2a\cos(\theta)\cdot(\cos\theta \vec u+\sin\theta \vec v)$ such that $b\gt0$. This is equivalent to the condition that the angle between $A$ and $B$ is acute.

share|cite|improve this answer

If you only have $\|a-b\|=\|a\|$, (it is a convention to express vectors using small letters), as @did mentioned, the angle cannot be determined.

The angle can be computed using the cosine formula as below. $$\cos\theta=\frac{\|a\|^2+\|b\|^2-\|a-b\|^2}{2\|a\|\|b\|}=\frac{\|b\|}{2\|a\|}$$ In order to compute the angle, you need also know $\|a\|$ and $\|b\|$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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