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I am working on a problem which requires me to find certain values of the components of vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^4$ such that the angle between them is $\pi/3$ If my understanding is correct then the answer should be the values for which $\arccos{((u\cdot v)/(|u| \cdot |v|))} = \pi/3$

But I am having a little trouble understanding what an angle means in higher dimensional spaces such as $\mathbb{R}^4$. So what is the definition of an angle in $n$ dimensions? Am I misunderstanding the problem?

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  • $\begingroup$ in any dimension, two vectors anchored at the origin form either a plane or a line. now you are back to finding the angle between two lines on a plane. $\endgroup$
    – abel
    Nov 28, 2014 at 15:49
  • $\begingroup$ See also the related question math.stackexchange.com/q/583066/9754 $\endgroup$
    – davidlowryduda
    Dec 4, 2016 at 19:07

1 Answer 1

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In an inner product space (like $\mathbb R^4$ or $\mathbb R^n$ with the standard inner product), the Cauchy-Schwarz inequality says that $|u\cdot v|\le \|u\|\|v\|$. Thus $-1\le \frac {u\cdot v}{\|u\|\|v\|}\le 1$ for all non-zero vectors. That means that $\arccos \frac {u\cdot v}{\|u\|\|v\|} $ is defined, and that is defined to be the angle between the vectors.

To understand the reason for this definition, one only needs to look at the law of cosines in dimension two.

Now, the thing is that the angle between two vectors is a quantity completely determined by the span of the two vectors, which is a plane. So, it doesn't matter what the dimension of the ambient space is, once you have two vectors, they determine a plane and the angle lives there. You can also define the angle geometrically rather than algebraically. The nice thing in an inner product space is that the algebraic definition is very simple.

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  • $\begingroup$ Did you mean arccos or arctan $\endgroup$ Nov 28, 2014 at 20:18
  • $\begingroup$ thanks @KarthikUpadhya , updated. $\endgroup$ Nov 28, 2014 at 20:59

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