$s$, $p$ 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.
$s$, $p$ 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.
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.
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: