Find s in terms of p and c for certain equations $s$, $p$ and $c$ are vectors; I need to find $s$ in terms of the other two for:
(1) $| s - c | = 1$
(2) $s = \lambda p$ ( $\lambda$ is a constant )
How can I use the constant $\alpha = (p \cdot c)^2 - p^2 * (c^2 - 1)$?
There may be no solution.
 A: Just as in the other problem you posted, since $s=\lambda\cdot p$, if you know $\lambda$ then you know $s$. If $p=0$, then there is a solution if and only if $|c|=1$, in which case the solution is $s=0$.
Assume $p\neq 0$. 
Plugging $s=\lambda p$ into the first equation, you know that
$$|\lambda p - c| = 1.$$
Since $|v|^2 = v\cdot v$ for any vector $v$, this gives
$$1 = (\lambda p - c)\cdot (\lambda p - c) = \lambda^2 (p\cdot p) - 2\lambda(p\cdot c) + (c\cdot c),$$
or equivalently, that
$$(p\cdot p)\lambda^2 - 2(p\cdot c)\lambda + \bigl( (c\cdot c)-1\bigr) = 0.$$
This is a quadratic equation in $\lambda$; if the discriminant
$$4(p\cdot c)^2 - 4(p\cdot p)\bigl((c\cdot c)-1\bigr) = 4\alpha$$
(with $\alpha$ the correct version of what you write above; see note at the end) is negative, there are no solutions. If $4\alpha$ is nonnegative,
then solving for $\lambda$ gives that
$$\lambda = \frac{2(p\cdot c) + \sqrt{4(p\cdot c)^2 - 4(p\cdot p)\bigl((c\cdot c)-1\bigr)}}{2(p\cdot p)}$$
or 
$$\lambda = \frac{2(p\cdot c) - \sqrt{4(p\cdot c)^2 - 4(p\cdot p)\bigl((c\cdot c) - 1\bigr)}}{2(p\cdot p)}.$$
which yields (up to) two possible solutions for $\lambda$, hence up to two values for $s$. 
The constant $\alpha$ in your statement is a rather bad attempt at describing (one fourth of) the discriminant. $p^2$ should be $p\cdot p$ and $c^2$ should be $c\cdot c$. Using
$$\alpha = (p\cdot c)^2 - (p\cdot p)\bigl((c\cdot c) - 1\bigr)$$
and simplifying, we can write it as:


*

*If $p=0$, then no solutions if $|c|\neq 1$, and $s=0$ is the unique solution if $|c|=1$. 

*If $p\neq 0$ and $\alpha\lt 0$, then no solutions;

*If $p\neq 0$ and $\alpha\geq 0$, then the solutions are given by
$$\lambda = \frac{(p\cdot c) + \sqrt{\alpha}}{p\cdot p},\qquad\text{and}\qquad
\lambda = \frac{(p\cdot c) - \sqrt{\alpha}}{p\cdot p}.$$

A: The question is not entirely clear.  I will try to make a clear question from it, using more or less your notation.  It may not be the question you intended to ask!
Let the vectors $p$ and $c$ be given.  Find a constant $\lambda$ such that if $s=\lambda p$, then $|s-c|=1$.
Solution: For any vector $v$, $|v|=\sqrt{v\cdot v}$.  Since $s=\lambda p$, we want
$$(\lambda p -c)\cdot(\lambda p -c)=1$$
Expanding the dot product according to the usual rules, we obtain
$$\lambda^2 (p\cdot p) -2\lambda (p\cdot c) +c\cdot c-1=0$$
The above equation is a quadratic equation in $\lambda$ (unless $p$ is the zero-vector).
Solve for $\lambda$ using the Quadratic Formula.  Note that there may not be a (real) solution. 
