1. Find infinitely many non-proportional functions f(x) in C[0,1] which form angle Pi/3 with the function x2.

The condition of the problem means that

<f(x),x2>/(||f|| ||x2||)=1/2.

Therefore

<f(x),x2>=||f(x)||/(2sqrt(5)).

Thus

<f(x),x2>2 = ||f(x)||2/20. (***)
Take f(x)=xn+a where n is an arbitrary positive integer and a is a parameter. Then

<f(x),x2>2 = int from 0 to 1 (x2(xn+a))

which is

(1/(n+3)+a/3)2 .


||f(x)||2=1/(2n+1)+2a/(n+1)+a2


Now we can plug it in (***) and solve for a.

> A:=solve((1/(n+3)+a/3)^2=(1/(2*n+1)+2*a/(n+1)+a^2)/20,a); 

                                                                  2 1/2
                 60            9             n (226 n + 179 + 11 n )
      A := - ---------- + ---------- + 3/11 ----------------------------,
             11 (n + 3)   11 (n + 1)                                 1/2
                                            (n + 3) (n + 1) (2 n + 1)

                                                                 2 1/2
                60            9             n (226 n + 179 + 11 n )
          - ---------- + ---------- - 3/11 ----------------------------
            11 (n + 3)   11 (n + 1)                                 1/2
                                           (n + 3) (n + 1) (2 n + 1)
This gives us an expression of a in terms of n. Thus for every natural n we can find a such that the angle between xn+a and x2 is Pi/3. For example if n=2 then

> n:=2:print(A);
- 9/11 + 2/275 6751/2 51/2 , - 9/11 - 2/275 6751/2 51/2


Problem 2. Given the 3-vector A=(1,2,3).
Find two non-zero 3-vectors B and C such that A is orthogonal to B and C and B is orthogonal C.

We need to find B and C such that 
 
					   B*A=0, 
					   C*A=0, 
					   C*B=0.
We can find B by inspection: B=(3,0,-1). Clearly B is orthogonal to A. Now we need to find C=(x,y,z) such that C*A=C*B=0.

This leads to the following system of equations: 
 
					x+2y+3z=0 
					3x -  z=0
The augmented matrix is the following:

> with(linalg):A:=matrix([[1,2,3,0],[3,0,-1,0]]);
A := [ 1 2 3 0 ]
[ 3 0 -1 0 ]
The Gauss-Jordan procedure gives the following:

> gaussjord(A);
[ 1 0 -1/3 0 ]
[ 0 1 5/3 0 ]
Thus z is a free variable and x and y are leading variables. The general solution is: 
 
					   x=1/3z 
					   y=-5/3z 
					   z=z
Taking z=3 gives C=(1,-5,3). This C is orthogonal to our A and B.

Answer: (3,0,-1), (1,-5, 3).