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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.

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Every acute angle is possible. Draw a picture in 2D. –  Did Nov 8 '12 at 7:11

2 Answers 2

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.

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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 $\|b\|$.

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