How can I reflect a point with respect to a plane?


I have three points (0,0,2), (4,0,0) and (0,8,0). And I have a point (x,y,z). From these, I want derive a composite transformation matrix. How can I do that?

  • 2
    \$\begingroup\$ What have you tried? What problems did you run into? And lastly: smells like homework, is it? \$\endgroup\$
    – Eric
    Commented Nov 12, 2012 at 14:30
  • \$\begingroup\$ @Eric I donot know how to solve it ? \$\endgroup\$
    – gveaf
    Commented Nov 12, 2012 at 14:44
  • 1
    \$\begingroup\$ @user1796528 welcome. On this site, those kind of questions normally require you to have already some work done. This is because this is not a place to just wait for someone answer any and all of your questions. You have to show work \$\endgroup\$ Commented Nov 12, 2012 at 17:02
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    \$\begingroup\$ From the way your question is worded, it sounds like your first three points identify your plane and the fourth is the one you want to reflect about that plane. Is that correct? \$\endgroup\$
    – user1430
    Commented Nov 12, 2012 at 18:16

2 Answers 2


Given the following:

The calculation steps for the augmented reflection matrix are as follows:

1) Calculate the 3x3 reflection matrix R0 for a plane with the same normal vector, but which lies at the origin

Ro = I - 2 NNT

2) Augment the reflection matrix to create the augmented reflection matrix RA

3) Calculate the final augmented reflection matrix by first applying the translation A, mirroring the point by RA, then applying the reverse of the translation A-1

R = ARAA-1


Three points, A, B and C define a plane. The lines AB and AC are both on the plane, so their normalised cross product, perpendicular to both lines, is the plane's normal n:

n = normalise( (B - A) ⨯ (C - A) )

The equation for the plane is given by: n · x + d = 0, where -d is the displacement between the plane and the origin in the normal direction, i.e.: d = -n · A.

An arbitrary point x can be described as the sum of two vectors, u pointing from plane to point in the normal direction, v being the other component:

u = (x · n + d) n

v = x - u

The mirror image of x can then be calculated by flipping the u-component: x' = -u + v. Now that x' is known, capturing the calculation in a transformation matrix is trivial: solve T x = x' for T.

  • \$\begingroup\$ The question asks for a matrix solution. Your explanation is solid, just not quite what the asker is looking for. :) \$\endgroup\$ Commented Nov 13, 2012 at 9:32
  • \$\begingroup\$ Good note. The transformation matrix is easily found from here, but that's rather moot if you have calculated x' already. There are of course shortcuts and simplifications you could apply, but in my opinion, going into detail there calls for proper mathematical formatting. \$\endgroup\$ Commented Nov 13, 2012 at 11:47

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