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This seems to me as a very simple and basic question, though I'm having trouble with it.

The Problem

Given a sphere $K\in\mathbb{R}^n$ with radius $r\in\mathbb{R}$ and center $\vec{c}\in\mathbb{R}^n$: $K:||\vec{x}-\vec{c}||_2^2=r^2$

And a given line along $\vec{n}\in\mathbb{R}^n$, $||\vec{n}||_2=1$ through $\vec{x}\in\mathbb{R}^n$: $g: \vec{x}+\vec{n}d$

Substituting the line into the equation of $K$ gives $||\vec{x}+\vec{n}d-\vec{c}||_2^2=r^2$ which results in the quadratic equation $\sum_{i=1}^{n}(x_i+n_id-c_i)^2-r^2=0$

Within my problem, it is ensured, that the supporting vector $\vec{x}$ of the line is always within the sphere. With my understanding, this simplifies the solutions of the quadratic equation in such a way, that there are always two solutions for $d$.

The Question

And this is the point, where I got stuck. How can this be rewritten either by using the quadratic formula ($ax^2+bx+c=0 \Rightarrow x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a^2}$) or as a system of equations?

Any help is appreciated. Thank you.

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    Oh, thank you for this remark, @Mariano. I've deleted that tag.2011-06-21

1 Answers 1

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$\sum_{i=1}^{n}(x_i+n_id-c_i)^2-r^2=0$

$d^2\sum_{i=1}^{n}n_i^2+2d\sum_{i=1}^{n}n_i(x_i-c_i)+\sum_{i=1}^{n}(x_i-c_i)^2-r^2=0$

Now you can solve for $d$.

If you're familiar with dot product the whole thing can be simplified. $(\vec nd+\vec x- \vec c).(\vec nd+\vec x- \vec c)=r^2$

$d^2 \vec n.\vec n + 2d\vec n.(\vec x-\vec c)+(\vec x-\vec c).(\vec x-\vec c)=r^2$

$d^2 \|\vec n\|^2 + 2d\vec n.(\vec x-\vec c)+\|\vec x-\vec c\|^2=r^2$

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    Thanks a lot. It was easy, as I thought.2011-06-22