# Vectors exercise

1. Sep 25, 2004

### AronH

I was trying to learn something about vectors in greater dimensions then only 3. I got this exercise and I couldn't get to it. Can anyone help me?
"Let n and a be vectors in R n , and let P = {× Î R n: n . (x-a) =0}.
Find an equation in x1,x2,...xn such that x = (x1,x2,x3...,xn) Î P if and only if the coordinates of x satisfy the equation." (Î means "be part of")
Thas is it.
Thank you all.

Aron

2. Sep 25, 2004

### HallsofIvy

Did this exercise really use n both as a vector and as the dimension???

I'm going to use v in place of the vector "n".

Write v= (v1, v2, v2,..., vn) and a as (a1, a2, a3,...,an).

Then x-a= (x1-a1, x2-a2,x3- a3,...,xn-an) and

v.(x-a)= v1(x1-a1)+ v2(x2-a2)+ v3(x3-a3)+...+ vn(xn- an)

(Suppose this problem had been about R<sup>3</sup>? What would you have done then?)

3. Sep 26, 2004

### AronH

I'm going to use v in place of the vector "n".

Write v= (v1, v2, v2,..., vn) and a as (a1, a2, a3,...,an).

Then x-a= (x1-a1, x2-a2,x3- a3,...,xn-an) and

v.(x-a)= v1(x1-a1)+ v2(x2-a2)+ v3(x3-a3)+...+ vn(xn- an)

(Suppose this problem had been about R<sup>3</sup>? What would you have done then?)

First of all, thank you vey much for the help, and the problem really use n for both dimension and vector, not very good but that isn't very important.
About the problem, I thought in the expression for x to be a system like this:
x1=a1 + v2(x2-a2)/v1 + v3(x3-a3)/v1 + ... +vn(xn-an)/v1
x2=a2 + v1(x1-a1)/v2 + v3(x3-a3)/v2 + ... +vn(xn-an)/v2
.
.
.
xn=an + v1(x1-a1)/vn + v2(x2-a2)/vn + ... +vn-1(xn-1-an-1)/vn

Is it something like that?
Now for R <sup>3</sup> I think that it is a particular case of that, so it is going to be something like this:
x1=a1 + v2(x2-a2)/v1 + v3(x3-a3)/v1
x2=a2 + v1(x1-a1)/v2 + v3(x3-a3)/v2
x3=a3 + v1(x1-a1)/v3 + v2(x2-a2)/v3
for a better view i called, v1/v2=i;v1/v3=j;v2/v3=k; and also defined un=xn-1n ao we have,
x1=a1+u2/i+u3/j
x2=a2+u1*i+u3/k
x3=a3+u1*j+u2*k
What more I can get from this in R3?
Well, thank you again,

Aron
PS- My first propouse is to learn vectors and aply calculus to more than one variable to use it in physics general physics. Is it a good think to see or there are useless points and an easy and more effective away to get where I want? (I already know basic calculus, derivation, integration and know a little about differential equation, but I always get some problems to aply this to physics, more especific the part that need some geometry.

4. Sep 26, 2004

### mathwonk

There are two ways to slip up on this problem, either not seeing how to analyze the problem, or not knowing the formula for a dot product. I am guessing it was the former.
I.e. the statement was that x belongs to the set P if and only if n.(x-a) = 0. Since the dot product of two vectors measures the product of their lengths times the cosine of the angle between them, the dot product being zero means that either one of the vectors has length zero, or the angle between them is pi/2. Thus the set consists of all points on the hyperplane passing through the point a, and such that this hyperplane is perpendicular to the vector n. (That is why n was used, since the vector is "normal" to, i.e. perpendicular to, the hyperplane. Probably in the original it was bold, or had an arrow over it to designate a vector, as opposed to the number n. Here the letter is sued for the point thought of as a "point", i.e. the head of the vector, and the letter n is used for the one thought of as a direction "vector", i.e. an arrow joining two points.)

Now you were asked to find an "equation" describing your set, which means something with an "equals" sign in it. But if you look at it as it was given, you see it already has an equals sign in it: namely n.(x-a) = 0. So that is already an equation describing the set. Thus all you were supposed to do was transform this same equation into an equation involving coordinates of the given quantities in the equation. So all you needed to know, is how to write out a dot product in coordinates.

i.e. if we write out n in coordinates, and also write out x-a in coordinates we get the equations:
0 = v.(x-a)= v1(x1-a1)+ v2(x2-a2)+ v3(x3-a3)+...+ vn(xn- an), as Halls of Ivy said.

that is already the equation you want.

Now a third potential problem, is actually solving the equation, and you do not seem to have this quite nailed. E.g. your first equation has minus sign errors. I.e.

x1=a1 + v2(x2-a2)/v1 + v3(x3-a3)/v1 + ... +vn(xn-an)/v1,

should be:

x1=a1 - v2(x2-a2)/v1 - v3(x3-a3)/v1 - ... -vn(xn-an)/v1.

Now provided v1 is not zero, this is the whole story. There is no system, just one equation. I.e. for any values at all of x2,....,xn, you get a solution to your original equation by choosing x1 to satisfy this one equation. It looks as if you should review solving systems of linear equations in several unknowns, beginning with systems of just one equation..

5. Sep 26, 2004

### AronH

Thank you Mathwonk,

I got the points that you talked about.
And sorry guys for the last post I was in a hurry so there were a lot of wrong words and signals too. (I had that minuses in my paper here) Also English isn't my native language so.
Thank you again,

Aron