I'm working on a programming project. In this project, a ray is fired from a point in 4-space. I need to find the distance from this ray to a number of other points in 4-space.

I attempted to solve this myself, but failed. The ray, $r$, can be written as $a + bt$, where $a$ and $b$ are vectors and t is the distance along the ray. The distance from this ray to a point $p$ can be written as $\lvert p - (a + bt)\rvert$. Dropping the square root (minimizing squared distance will also minimize distance), this is

$$((p_x - a_x) - b_xt) ^ 2 + \cdots$$

Differentiating with respect to $t$ yields

$$2(p_x - a_x - b_xt)(-b_x) + \cdots = 0$$


$$-p_xb_x + a_xb_x + b_x^2t + \cdots = 0$$

Solve for $t$…

$$t(b_x^2 + b_y^2 + \cdots) = p_xb_x - a_xb_x + \cdots \\ t = (p_xb_x - a_xb_x + \cdots) / (b_x^2 + \cdots)$$

In my case, $b$ is a unit vector, so the bottom simplifies.

$$t = (p_xb_x - a_xb_x + \cdots) \\ t = b\cdot(p - a)$$

So the distance should be

$$\lvert p - a - b(b \cdot (p - a))\rvert = d$$

Is that correct? I'm not sure if the problems I'm having are a result of my distance formula or are a problem with other aspects of my program.


1 Answer 1


Your formula looks correct for $\lVert b\rVert=1$ as you stated. You take the difference vector $(p-a)$ and from that difference vector you determine the distance in direction of $b$, whgich is $b\cdot (p-a)$, then substract that component parallel to $b$ in $(p-a)-(b\cdot (p-a))b$. The result is the component of $(p-a)$ orthogonal to $b$, which is exactly what you need.

  • $\begingroup$ It is correct. I checked with known int rather than my Monte-Carlo simulation. It's just ugly, so I need to redesignth e program. $\endgroup$ Apr 21, 2014 at 15:47

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