# Thread: Analysis question from Baby Rudin (PMA), Chapter 1

1. ## Analysis question from Baby Rudin (PMA), Chapter 1

This is number 19 from Chapter 1 of "Principles of Mathematical Analysis." It's really confusing to me. Here we go:

Suppose a belongs to R^k, and b belongs to R^k. (Where R^k is a Euclidean Space) Find c in R^k and r>0 such that
|x - a| = 2|x - b|
if and only if |x - c| = r.
Solution: 3c = 4b - a, 3r = 2|b - a|.

I've put the "vectors" in bold type. Also the "solution" is given in the book. I'm not sure how we're supposed to use it... whether or not we simply plug it in and show that it works, or whatever...

2. Originally Posted by iamthemanyes
Suppose a belongs to R^k, and b belongs to R^k. (Where R^k is a Euclidean Space) Find c in R^k and r>0 such that
|x - a| = 2|x - b|
if and only if |x - c| = r.
Solution: 3c = 4b - a, 3r = 2|b - a|.
Use the inner product $\displaystyle \langle.,.\rangle$ in $\displaystyle \mathbb{R}^n$ to write $\displaystyle |\mathbf{x}|^2 = \langle\mathbf{x},\mathbf{x}\rangle$. Square both sides of the equation $\displaystyle |\mathbf{x}-\mathbf{a}| = 2|\mathbf{x}-\mathbf{b}|$, getting $\displaystyle \langle\mathbf{x}-\mathbf{a},\mathbf{x}-\mathbf{a}\rangle = 4\langle\mathbf{x}-\mathbf{b},\mathbf{x}-\mathbf{b}\rangle$. Multiply out and rearrange the terms to get this into the form $\displaystyle \langle\mathbf{x}-\tfrac13(4\mathbf{b}-\mathbf{a}),\mathbf{x}-\tfrac13(4\mathbf{b}-\mathbf{a})\rangle = \ldots$.

(It's like completing the square, but using inner products instead of ordinary algebra.)