Length of sum of high-dimensional 3 vectors in terms of their lengths and inner products I have 3 vectors in a high-dimensional euclidean space.  I would like to express the length of their sum in terms of their lengths and inner products. For two vectors, I can do it with the law of cosines:
$c^2 = a^2 + b^2 - 2ab\cos\theta$
which in my context translates to:
$||c||^2 =||a||^2+||b||^2 -2 a \cdot b$
(where c = a+b and all are vectors, I project them into the span of a,b to get the triangle and then apply the law of cosines from trig)
How do I do this for 3 vectors, d=a+b+c?
I'll keep working on this and answer my own question if I come up with a solution.

Hint: Use associativity:  $$\overline{a}+\overline{b}+\overline{c}=(\overline{a}+\overline{b})+\overline{c}$$
This yields the answer since you already know how to add two vectors.  As a check, notice that the answer has to be the same if I permute $a,b,c$ since addition is commutative.
Edit:
Say you have vectors $a,b,c$.  Then $$\|a+b+c\|=\|(a+b)\|^2+\|c\|^2-2(a+b)\cdot c$$
$$=\|(a+b)\|^2+\|c\|^2 -2a\cdot c-2b\cdot c$$
$$=\|a\|^2+\|b\|^2+\|c\|^2 -2\left( a\cdot c+b\cdot c+a\cdot b\right)$$
