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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

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."

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
plane:

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

and:

a + b + c = 1

since I is also on P.

Substituting:

(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'.

If:

M = (p,q,r)

then:

p + q + r = 1

and:

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

Thus:

(-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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