Here's a summary of Dimitris Andreou's link.

Remember sum of i-th powers, where i=1,2,..,k. This reduces the problem to solving the system of equations

a_{1} + a_{2} + ... + a_{k} = b_{1}

a_{1}^{2} + a_{2}^{2} + ... + a_{k}^{2} = b_{2}

...

a_{1}^{k} + a_{2}^{k} + ... + a_{k}^{k} = b_{k}

Using Newton's identities, knowing b_{i} allows to compute

c_{1} = a_{1} + a_{2} + ... a_{k}

c_{2} = a_{1}a_{2} + a_{1}a_{3} + ... + a_{k-1}a_{k}

...

c_{k} = a_{1}a_{2} ... a_{k}

If you expand the polynomial (x-a_{1})...(x-a_{k}) the coefficients will be exactly c_{1}, ..., c_{k} - see Viète's formulas. Since every polynomial factors uniquely (ring of polynomials is an Euclidean domain), this means a_{i} are uniquely determined, up to permutation.

This ends a proof that remembering powers is enough to recover the numbers. For constant k, this is a good approach.

However, when k is varying, the direct approach of computing c_{1},...,c_{k} is prohibitely expensive, since e.g. c_{k} is the product of all missing numbers, magnitude n!/(n-k)!. To overcome this, perform computations in Z_{q} field, where q is a prime such that n <= q < 2n - it exists by Bertrand's postulate. The proof doesn't need to be changed, since the formulas still hold, and factorization of polynomials is still unique. You also need an algorithm for factorization over finite fields, for example the one by Berlekamp or Cantor-Zassenhaus.

High level pseudocode for constant k:

- Compute i-th powers of given numbers
- Subtract to get sums of i-th powers of unknown numbers. Call the sums b
_{i}.
- Use Newton's identities to compute coefficients from b
_{i}; call them c_{i}. Basically, c_{1} = b_{1}; c_{2} = (c_{1}b_{1} - b_{2})/2; see Wikipedia for exact formulas
- Factor the polynomial x
^{k}-c_{1}x^{k-1} + ... + c_{k}.
- The roots of the polynomial are the needed numbers a
_{1}, ..., a_{k}.

For varying k, find a prime n <= q < 2n using e.g. Miller-Rabin, and perform the steps with all numbers reduced modulo q.

EDIT: The previous version of this answer stated that instead of Z_{q}, where q is prime, it is possible to use a finite field of characteristic 2 (q=2^(log n)). This is not the case, since Newton's formulas require division by numbers up to k.

It's not clear from your question, but it sounds like you have a line defined by two points, and you want to find the distance between that line and a third point? As you've seen, you can't uniquely determine A,B,C and D from just two points -- there are an infinite number of planes that contain that line.

If x_1 and x_2 are the 3-d vectors representing the points defining the line, and x_0 is the point
you're trying to find the distance to, the formula is:

```
d = (|(x_2-x_1)x(x_1-x_0)|)/(|x_2-x_1|)
```

(In this formula, 'x' denotes the 3-d cross product, '-' denotes vector subtraction, and |v| represents the length of vector v.)

Point-Line Distance--3-dimensional

## Best Solution

Given point A and a line, pick two different points on the line (B and C). Calculate the area of the triangle ABC with use of Heron's formula. Multiply the area by 2 and divide it by length of [BC]. You have the result you needed.