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Baby Rudin exercise 1.19

  • Thread starter lol_nl
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  • #1
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Homework Statement


Suppose [itex] a \in R^{k},b \in R^{k}[/itex]. Find [itex] c \in R^{k}[/itex] and [itex] r > 0[/itex] such that
|x-a| = 2|x-b|
if and only if |x-c| = r.

Homework Equations


Solution:
3c = 4b - a, 3r = 2|b-a|

The Attempt at a Solution


Technically, I can't seem to find any way to express |x-c| in terms of |x-a| or |x-b| except using the triangle inequality.
The idea of the exercise is a complete mystery to me. What is the use of proving this theorem?
 
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Answers and Replies

  • #2
My advice is to start with a sketch in 2d. What is the set of x for |x-a|=const?
 
  • #3
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It's a circle. My intuition tells that the set that satisfies the first equation is a hyperbola: the difference of two distances is a constant (namely 0). Only the factor two seems misleading...
 
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  • #4
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What you have here are two arbitrary points in space, a and b. x is the third point, thus forming a triangle, in which the length of side ax is twice the length of the side bx. The length of side ab is fixed. How many such triangles can you form in 2D? 3D?
 
  • #5
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What you have here are two arbitrary points in space, a and b. x is the third point, thus forming a triangle, in which the length of side ax is twice the length of the side bx. The length of side ab is fixed. How many such triangles can you form in 2D? 3D?
In 2 dimensions the answer seems to be 2: with the length of all three sides determined, any two such triangles must be congruent, and also share the side ab. In three dimensions, there is an entire circle.
However, letting |x-a| vary over all positive values, one gets a curve in two dimensions and probably a surface in three dimensions. My guess was that the curve is a hyperbola, but I haven't managed to figure out whether that is true.
 
  • #6
6,054
390
However, letting |x-a| vary over all positive values, one gets a curve in two dimensions and probably a surface in three dimensions.
Can you really have x vary here? Work this out in 2D: given a fixed side ab, find all the locus of all x that must satisfy ax = 2bx. Hint: project x onto ab, and call that point c.
 
  • #7
How about for starters, you put A=(0,0) and B=(b,0)
Then you calculate e1=|x-A| and e2=|x-B|
Then you let e1=e2, simplify and examine what you get.
This would give you an intuition about the nature of the problem.
You still need to carry out the calculation in k-dim and general case, though.
 
  • #8
SammyS
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Homework Statement


Suppose [itex] a \in R^{k},b \in R^{k}[/itex]. Find [itex] c \in R^{k}[/itex] and [itex] r > 0[/itex] such that
|x-a| = 2|x-b|
if and only if |x-c| = r.

Homework Equations


Solution:
3c = 4b - a, 3r = 2|b-a|

The Attempt at a Solution


Technically, I can't seem to find any way to express |x-c| in terms of |x-a| or |x-b| except using the triangle inequality.
The idea of the exercise is a complete mystery to me. What is the use of proving this theorem?
Look at a, b, c, and x as k-dimensional position vectors.

Squaring |x-a| = 2|x-b|, gives (x-a)∙(x-a) = 4(x-b)∙(x-b) where the ∙ symbol denotes the inner product. (That's the scalar product in 3.)

This can be expanded as [itex]\displaystyle \textbf{x}\cdot\textbf{x}-2\textbf{x}\cdot\textbf{a}+\textbf{a}\cdot\textbf{a}=4\left(\textbf{x}\cdot\textbf{x}-2\textbf{x}\cdot\textbf{b}+\textbf{b}\cdot\textbf{b}\right)\ .[/itex]

This ca be written as [itex]\displaystyle \textbf{x}^2-2\textbf{x}\cdot\textbf{a}+\textbf{a}^2=4\textbf{x}^2-8\textbf{x}\cdot\textbf{b}+4\textbf{b}^2\,,[/itex] if we use the convention that [itex]\displaystyle \textbf{u}^2=\textbf{u}\cdot\textbf{u}\,,[/itex] for any k-vector, u. Now, collect terms and complete the square.
 

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