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I would like to know if it's possible, given the vector equation of a line and the coordinates of a point, whether it's possible to reflect the point by the line.

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    $\begingroup$ Yes. I'm not sure what tools you have at your disposal but it goes something like this: If the point is in the line, then its reflection is the point itself. If the point isn't in the line, just take the line perpendicular to the given line which intersects the point. The reflection will be the only OTHER point on the perpendicular line which is at the same distance to the initial line as the first point is. $\endgroup$
    – Git Gud
    Commented Jan 18, 2013 at 16:58
  • $\begingroup$ You might also use this. $\endgroup$ Commented Jan 18, 2013 at 17:02
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    $\begingroup$ The terminology is slightly ironic in that "reflecting" a point by a line in $\mathbb{R}^3$ is actually a half-full rotation of $\mathbb{R}^3$ around that line (axis). $\endgroup$
    – hardmath
    Commented Jan 18, 2013 at 17:20

2 Answers 2

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Given a line $\overline r+t\cdot\overline{v}$ and a point to be reflected $\overline p$, we must first find the closest point on the line to $\overline{p}$. Let $t_0$ correspond to that point. Its value is:

$$t_0 = \dfrac{\overline{p}\cdot\overline{v}-\overline{r}\cdot\overline{v}}{\overline{v}\cdot\overline{v}}$$

The intersection on the line is hence $\overline{s} = \overline{r}+t_0\cdot\overline{v}$, and the reflected point $\overline{p}_2 = 2\overline{s}-\overline{p}$.

About the solution: one has to find a line perpendicular to the given line, but also through the point to be reflected. A vector $\overline{d}$ perpendicular to $\overline{v}$ is given by the fact that if $\overline{d}\bot \overline{v}$, then $\overline{d}\cdot\overline{v}=0$. On the other hand, we must arrive to the same point through two different routes, $\overline{r}+t\cdot\overline{v}$ and $\overline{p}+\overline{d}$ so we set them equal. We now have a system of two vector equations to solve.

With all due respect, I don't think the answer Ross provided here was quite right. Using my notation, he suggested, that $t_0 = -\dfrac{\overline{p}\cdot\overline{r}}{\overline{p}\cdot\overline{v}}$, which is not the right solution.

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    $\begingroup$ This is the correct solution. Ross' is incorrect. $\endgroup$ Commented Aug 15, 2018 at 7:20
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Yes, if by reflect you mean to draw a perpendicular from the point to the line and continue it the same distance on the other side. If your line is $(p_x,p_y,p_z)+t(q_x,q_y,q_z)$ (is this what you mean by vector equation?) and the point is $(r_x,r_y,r_z)$ the point on the line where the perpendicular hits can be found by the condition that the dot product with the direction vector is zero. We want to find $t$ such that $r_x(p_x+tq_x)+r_y(p_y+tq_y)+r_z(p_z+tq_z)=0$. This is a linear equation that can be solved $t=-\frac {\vec r \cdot \vec p}{\vec r \cdot \vec q}$ The perpendicular point is then point is then $\vec s=\vec p+t \vec q$ and the reflected point is then $2\vec s-\vec r$.

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    $\begingroup$ This is incorrect. You have used the equation for finding the t value for where the line cuts the plane through the origin perpendicular to r. $\endgroup$ Commented Aug 15, 2018 at 7:26

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