Reflecting a point by a line in $\mathbb R^3$ 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.
 A: 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.
A: 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 $2\vec s-\vec r$.
