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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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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. – Git Gud Jan 18 '13 at 16:58
You might also use this. – David Mitra Jan 18 '13 at 17:02
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). – hardmath Jan 18 '13 at 17:20
up vote 3 down vote accepted

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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