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Let $x$ and $y$ be two vectors and $A$ the angle between them. Then we have the scalar product $$x\cdot y = \|x\|\|y\| \cos A$$

Let $x$, $y$ and $z$ be three vectors; $A$ angle between $x$ and $y$; $B$ angle between $x$ and $z$; and $C$ angle between $y$ and $z$. What is the value of the scalar product for the three vectors? Generalization: What is the value of the scalar product for $N$ vectors in $n$-dimensional space?

In 2-dimensional space we define a symmetric bilinear form for scalar product. In n-dimensional space can we define a symmetric multilinear form for N vectors?

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  • $\begingroup$ Welcome to MSE! For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – KittyL Jan 3 '16 at 18:28
  • $\begingroup$ I think this reference might help you :Vectors in n-dimensional Space The Cauchy-Schwarz and Triangle Inequalities It mentions: The algebraic definitions of vector addition, multiplication by scalar, and the dot product, extend to n-dimensional space in the obvious way. (The geometric interpretations of these operations (e.g., dot product formula in terms of angles and magnitudes) make only sense in $R_2$ and $R_3$ and do not extend to n-dimensional spaces. Also, the cross product, as defined in class, makes only sense in $R_3$.) $\endgroup$ – Mufasa Jan 3 '16 at 18:30
  • $\begingroup$ It's a good idea! Can we just do it the obvious way -- $(a, b) \cdot (c, d) \cdot (e, f) = ace + bdf$? Would you want to value to be related to all of the angles $A, B, C$? i.e. $x \cdot y \cdot z = ||x||||y||||z||\cos(A)\cos(B)\cos(C)$ or something similar? $\endgroup$ – Eli Rose Jan 3 '16 at 18:30
  • $\begingroup$ You can certainly think of symmetric $k$-linear forms as generalizations of the dot product, but they don't necessarily have any nice geometric interpretation in terms of lengths and angles the way that the dot product does. For instance, the "obvious" generalization of the dot product given by $T(a,b,c) = \sum_i a_ib_ic_i$ is coordinate-dependent (and thus not very useful). $\endgroup$ – user137731 Jan 3 '16 at 18:35
  • $\begingroup$ @Mufasa: What your source says is true, but the OP's question is about a function that would take in $n$ vectors and give a scalar, which I don't think is mentioned there. (Also, clarifying more for myself than anyone else, it's still true that $x \cdot y = ||x||||y||\cos(\theta)$ when $x, y \in \mathbb{R}^n$ for $n > 3$, the author's point is just that it's hard for humans to think about $4$-dimensional angles). $\endgroup$ – Eli Rose Jan 3 '16 at 18:37
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The notion of scalar product is defined for 2 vectors, not only for 2-dimensional spaces but for n-dimensional spaces as well, as is the corresponding (for the geometric interpretation) notion of the angle, which also only takes two vectors.

A generelisation in linear algebra is a scalar product that is defined as a bilinear, symmetric (as you said) and positive definite (i.e. $\left<v,v\right>>0 $ for $v\neq0$) map $V\times V \rightarrow \mathbb{K}$ where $\mathbb{K}$ is either $\mathbb{C}$ or $\mathbb{R}$ and $V$ is a vector space.

Now, independent of the dimension of your space you can define multilinear maps which are a generalisation of bilinear maps and can also have properties like symmetry, but scalar product is a term that refers to maps that takes 2 arguments in all occasions I've encountered so far.

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  • $\begingroup$ What is a N-dimensional angle? How do we justify the 'projection' intuition to higher dimensional spaces? Trigonometry helps me only in 2D… $\endgroup$ – Atcold Jan 22 '18 at 7:07

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