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The Reflection of a Line

Date: 04/21/98 at 16:02:13
From: Jessica
Subject: Vector equation

Determine a vector equation of the REFLECTION of the line:
  r(vector) = (0,2,5) + t(-1,1,3) in the plane x + y + z = 1

Please help me. I have tried for two weeks, negating, switching, and 
basically fumbling the number around. Thank you.

Date: 04/24/98 at 15:59:41
From: Doctor Ujjwal
Subject: Re: Vector equation

Dear Jessica,

Thank you so much for such an interesting problem! This problem can be 
solved using matrices that 'transform' a given figure, that is, 
translate, rotate, scale or reflect a given geometric figure. But such 
a solution unfortunately makes little intuitive sense. So let's try a 
different approach.

Let's see if we can simplify the problem.

1. The reflection of a line must be a line.
2. A line can be defined by any two points on it. 
3. If we can find the reflection of one point, the same method can be
   used to reflect two or more points.

So the problem really boils down to:
   "Given a point, find its reflection about the given plane."

Let's now think about this simpler problem.

4. The line joining a point and its image (reflection) will be normal
   to the mirror (the given plane).
5. To find the normal to a plane we can take the cross product of any
   two vectors in that plane.
6. A vector in a plane can be constructed using any two points in 
   the plane.

Let's see what we can do with the simple observations made above.

Let L be the line  r = (0,2,5) + t(-1,1,3)
and P be the plane x + y + z = 1

We will use angular brackets <> for vectors and parenthesis () 
for points.

Step 1
Take an arbitrary point on the line L. For ease of calculation let 
t = 0. We get a point (0,2,5). Let's call it A.

Step 2
To construct the normal to the plane we need two vectors in it. And to 
construct two vectors, we need at least three points. Let's find them 
next. Again for ease of calculation use z = 0 and y = 0 in 
x + y + z = 1. This gives us a point (1,0,0). Let's call it X.
Similarly we can get points Y(0,1,0) and Z(1,0,0). Now we can 
get vectors:
     XY = (0,1,0) - (1,0,0) = <-1,1,0> and
     XZ = (0,0,1) - (1,0,0) = <-1,0,1>

The cross product of XY and XZ will give a vector N normal to the 

     N = <-1,1,0> X <-1,0,1> = <1,1,1>

[Note: Looks like we are on the right track. Did you see that the 
normal is 'equally inclined' with the coordinate axes? That is because 
the plane itself is so oriented. This can be seen from its equation].

Step 3
Now we have the point A and the normal N. We can now construct a line 
that joins point A to its reflection A' as:

     r = A + u.N     
       = (0,2,5) + u*<1,1,1>

where u is a parameter (sort of distance from A). The line segment AA' 
will be bisected by the 'mirror', the plane P. Let I(a,b,c) be the 
point of intersection of AA' and P. Thus:

    (a,b,c) = (0,2,5) + u*<1,1,1>

because I is on AA'. Then: 

     a = u; b = 2 + u and c = 5 + u


     a + b + c = 1 

since I is also on P.


    (u) + (2 + u) + (5 + u) = 1
                     3u + 7 = 1
                          u = -2 

where u is the 'distance' of I from A. Since the mirror is -2 units 
from point A; we only need to go -2 units further 'into the mirror' to 
reach the reflection A'!

So use u = -4 to find A':

    A' = (0,2,5) -4*<1,1,1>
       = (-4,-2,1)            

This is a point on the reflection of L.

Step 4
You can repeat the above steps for another point B on L. Find its 
reflection B' and then find the line A'B'. 

But you can take an interesting (and quicker) approach here too. Take 
a point M on L that is also on P. Remember, "when you touch the 
mirror, the reflection touches you!" So M will coincide with its 
reflection M'.


      M = (p,q,r)


      p + q + r = 1       


     (p,q,r) = (0,2,5) + t*<-1,1,3>

      (-t) + (2 + t) + (5 + 3t) = 1
                         3t + 7 = 1
                              t = -2

We find that:

      M = (2,0,-1) = M'

Given two points A' and M' can you find the 'reflection' line?

It was fun solving this problem. Hope you enjoy finishing it. If you 
have questions regarding any of the steps or formulations, do write 
to us.

Good luck!

-Doctor Ujjwal Rane,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
Associated Topics:
College Higher-Dimensional Geometry

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