L2-0: Geometry: Review or New, its all just
plane talk
10 Oct 2015
Professor Henry Higgins: All right, Eliza, say it again.
Eliza Doolittle: The rine in spine sties minely in the pline.
Higgins: [sighs] The *rain* in *Spain* stays *mainly* in
the *plain*.
Eliza: Didn't ah sy that?
Higgins: No, Eliza, you didn't "sy" that, you didn't even
"say" that. Now every night before you get into bed, where
you used to say your prayers, I want you to say "The rain in
Spain stays mainly in the plain" fifty times. You'll get much
further with the Lord if you learn not to offend His ears.
-- My Fair Lady
Movie trivia questions: My Fair Lady was an
adaptation of George Bernard Shaw’s play Pygmalion. What was the basic story of Pygmalion?
Another famous Victorian playwright wrote a play based on Pygmalion, that was itself
subsequently parodied as Pygmalion Re-Versed. Is it true that the matrix product Pygmalion ReVersed . Pygmalion is the identity matrix I?
Primary concepts in this quasi-review: lines, planes, vectors, located vectors, direction
vectors, parameters, parametric equations, linear combinations, normal vector, unit normal
vector.
We begin on a familiar plane: In the x,y plane of two dimensional space (often written as R2),
the equation of a line is usually something familiar like y = mx + b or maybe the slightly more
quixotic ax + by = c.
Be sure to note that when we use these two conventional forms, those two different b’s
represent different things. Important questions: If a line goes through the origin of R2,
how do these equations simplify? How many “dimensions” does a line in R2 represent? If
we are to enjoy any of the benefits of linear algebra, should this dimensionality of a line
hold for any higher dimensional space Rn? ? (Note: that idea may no longer be review).
While the x, y plane is nice (and familiar), it’s kind of a plain plane. How would we write the
equation of a line through the origin in our everyday three-dimensional world, which we now
call R3? Seems like x, y and z all ought to get in that equation somewhere. Maybe it would be
simpler to first write equations for lines that are only in the (x, y, 0) plane, the (x, 0, z) plane and
the (0, y, z) plane. What do you get when you try these?
This leads to the exciting question: How would we write
the equation for a line in any orientation in any Rn space?
How many dimensions does one of those lines have
3
2
y
1
0
-1
2.0
Plot of a line in the 3-space known as R3.
1.5
Can you deduce the equation of this particular line?
z 1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
x
Use the vector equation applet at http://www.walterfendt.de/m14e/line3d.htm to play with lines in R3. Be sure you know what the parameter 
(lambda) represents in this applet.
Practice with these concepts before continuing.
Parametrically speaking
We can define a line through the origin in terms of a vector direction (which we will call p0) and
a scalar, x = s p0. As the scalar (or parameter) s takes on all different real number values
(including 0), x traces out all points in the direction specified by p0.
So in order to formally define a line L in any space Rn, we need just two things:
a starting point on the line and a direction in which to go from that point.
We will use only vector notation,
illustrated in the figure at right:
Let the starting point on the line
be represented by the vector p0
and let an arbitrary point on the
line be represented by the vector
p. Both vectors p0 and p are
located (they have their tails
firmly planted at the origin).
But our line must represent all
points along a trip from p0
Vector from p to p0
Vector from p0 to p
Vector p0
Origin
Vector p
2
through p (or from p through p0) and beyond, so we need a vector that represents one of these
directions (say p0 to p). What is that direction vector v in terms of p0 and p?
Did you note that our diagram makes no specification of what Rn it represents? Vectors
are happy to live in any Rn we like! And lines in higher dimensional spaces are still just
lines!
Now all we need is a scalar to stretch and squeeze the direction vector, so that we can get all
points p on our little ol’ line.
We can write the vector equation of a line L as p = p0 + t v, where v is the direction vector
from p0 to p and t is a scalar. This is called the parametric form of the line (the parameter
is the scalar t). The vector v is sometimes called the displacement vector – even though it
might have nothing to do with the physics term ‘displacement’! As noted way back in L11, it is sometimes customary to unitize (normalize) the vector v.
How many scalar parameters did we need to write this vector equation? How many dimensions
does that make our line? Does your answer depend upon the space it is in? Does your answer
depend on whether the line goes through the origin or not? What does the term dimension of an
object in space mean in this context?
1
2
1
3
4


and p2    .
Example: Write the vector equation of the line in R through the points p1 
1
4

 
1
5
1 
 2
  , so the line is represented by
p

p

Our direction vector is 2
1
 3
 
 4
 x  1 1 
 y  1  2 
p  p0  tv or p        t   . Yep, that’s a line.
 z  1  3 
    
 w 1  4 
3
We have a hard time plotting 4-space: If you lop off the w component, dropping down to R3,
you can use ParametricPlot3D, where the parameter is t, so that we may see the beauty of such
a thing -- especially with a nice Manipulate[ ]:
Manipulate[
Show[
Graphics3D[{
{Green, Line[{p0-dp, p0+dp}]},
{Red, PointSize[.03],Point[p0 + dp t]}
}], AxesTrue, AxesLabel{x,y,z}
],
{{t,0},-1,1}]
3
0.0
x
0.5
1.0
1.5
2
y
2.0
1
0
-1
4
2
z
0
-2
We can also write the problem as a system of four
equations in four unknowns,
x  x0  at , y  y0  bt , z  z0  ct , w  w0  dt , where t is still a parameter to be varied at will.
These parametric scalar equations give exactly the same result as the much more elegant vector
equation.
To solve these equations, we simply plug in the given values for the starting vector p0
(equivalent to setting p0 = {x0, y0, z0, w0}), yielding these equations:
x  1  at , y  1  bt , z  1  ct , w  1  dt
and then we eliminate the parameter t. Our line is now symbolized in compact form by
x 1 y 1 z 1 w 1



.
a
b
c
d
4
x
 y
Any point p    is on this line when it satisfies these equations – and we can obtain
z
 
 w
such a point from the information we were given.
Find values a, b, c, d that solve this problem. Did you get a = 1, b = 2, c = 3 and d = 4?
Is this solution unique – if not, why not? Does the non-uniqueness of a solution bother
you? If so, why? If not, why not?
And the bottom line, in plain-speak: Lines are one dimensional beings, made from vectors which may themselves be elements of a higher dimensional space. The vectors may have more
dimensions, but all lines are one dimensional!
Plane talk
In R3, a plane may be specified by a scalar equation in
standard form ax  by  cz  d .
Here’s a Plot3D example, with the equation of
the plane shown as a PlotLabel.
This can also be written so that the right hand side is exactly 1:
a
b
c
x  y  z  1 or in
d
d
d
d a
b
 x  y . And those coefficients may be combined into
c c
c
something like z  f ( x, y )     x   y , if you like a little Greek. Practice plotting these more
than one z = f(x,y) plane in the same Plot3D statement.
 x0 
If we have a point  y0  on this plane, then we could also write a scalar equation
 z0 
functional form z  f ( x, y ) 
a( x  x0 )  b( y  y0 )  c( z  z0 )  0 (what happened to d?)
In the vector-speak of an arbitrary Rn, we need a point p0 and two direction vectors v1 and v2 to
determine the plane. Ah, but these direction vectors must not be parallel! Perhaps we get the
new vectors from two additional points on the plane, p1 and p2: v1  p1  p0 and v2  p2  p0 .
5
This is known as a three-point problem, a cornerstone of much geometry. And this is where
Linear Algebra starts looking like geometry.
We parameterize the plane by starting at p0 and moving some distance in the v1 direction and
some other distance in the v2 direction: p  p0  sv1  tv2 . We now need two parameters, s
and t: A line has one dimension, a plane has two.
A plane is a linear combination of two vectors (provided that the vectors
are linearly independent). And those vectors can be in any Rn.
Say it this way:
There’s that linear combination again. You should note that all vectors p in the plane must
satisfy
p  p0  sv1  tv2
Restatement: The vector equation of a line is
p  p0  tv ; the vector equation of a plane is
p  p0  sv1  tv2 . So the vector equation of a 3-space is merely p  p0  rv1  sv2  tv3 .
What condition must the various v vectors satisfy? And why stop there?
Example
1  1
2
1  1 
3
Find the vector equation of the plane through   ,   and   .
1  1 
4
  
 
1  1 
5
Start by choosing one of the given vectors for p0 and find the two directions v1 and v2.
What happens if you choose a different p0? Are the solutions the same? Still bothered
by non-uniqueness?
Practice: Work through the 5 exercises in ‘finding the equation of a plane’ at
http://www.univie.ac.at/future.media/moe/galerie/geom2/geom2.html#eb . Write your
answers in parametric form.
x
 y
Now find the standard form equation of the plane. Start by locating an arbitrary point p   
z
 
 w
on the plane with four equations that come from the vector equation.
6
Another way to specifying the plane; and this one is normal (does that mean the prior is
abnormal?)
Suppose we have two vectors p1 and p2 in Rn that define two points in a plane; together they
determine a pair of displacement vectors (what are they?) that are also in the plane. We may
specify the plane by its normal vector n, requiring that n is perpendicular to the plane (and
therefore every vector in the plane).
We’ll use the fact that the dot product of any two perpendicular vectors is zero:
n( p2  p1 )  0
a 
Suppose the vector n   b  ; the equation of the plane through point
 c 
p1 = (x1, y1, z1) is just a( x  x1 )  b( y  y1 )  c( z  z1 )  0 .
Be sure you can show how does this equation comes from the dot product equation
above.
Example
1  1
2
1  1 
3
Use two of the points in the previous example (   ,   and   ) to find a vector normal to the
1  1 
4
  
 
1  1 
5
plane. Then use another pair of points to find another normal vector. Is it possible that there be
more than one n vector to a given plane? Consider only the direction indicated by the n vector;
magnitudes don’t matter in this question.
As stated above, a plane in R3 can be written in standard form: ax  by  cz  d . In this form the
a 
normal vector is just n   b  ; we sometimes want this to be a unit normal vector
 c 
a 
 b  , which is in the same direction as n. Oddly enough, finding a vector
nˆ 
2
2
2  
a b c  
c 
normal to a plane is not called “normalizing” the plane. Even more oddly, dividing a vector by
its own magnitude to create a vector of magnitude 1 is not called “unitizing” the vector (but it is
called normalizing the vector – even though the vector in question need not be normal to
anything of interest). Go figure.
1
7
For some primo-deluxe fun (and a really good way to look at planes in R3), play with the
Mathematica demo FromVectorToPlane.nbp. By now, you should see the relationship
between the vector equations and the algebraic equations for lines and planes.
Need more review? See http://www.maths.usyd.edu.au/u/MOW/vectors/vectors-12/v-12.html.
For an excellent visual demo of planes in space, see http://www.josechu.com/planes_in_3d/
Example
2
3


Find the equation of the plane through p0   4  with a normal vector n   5  .
 1
 2 
 x  p0 x   x  2 
The displacement vector is p  p0   y  p0 y    y  4  .
 z  p0 z   z  1 
The dot product equation of the plane gives us n p0  3( x  2)  5( y  4)  2( z  1)  0 .
We can now simplify this expression to obtain the standard equation of the plane: do it; then
verify that the normal vector n is really perpendicular to vectors in the plane. Do this for a
general case, using the vector expression for this plane and the appropriate dot product.
Practice
1
1 
 1




Find the standard form equation of a plane containing the three points p  1 , q   2  , r   2 
1
 0 
 1 
Take the conceptual practice quiz at
http://teachertech.rice.edu/Participants/nallen/lessons/quizzes/plpc.htm (* indicates a correct
answer, X indicates an incorrect answer)
Answer to movie trivia questions on page 1:
Pygmalion was a sculptor in Greek mythology; he created a sculpture that came to life.
The other playwright was W. S. Gilbert of Gilbert and Sullivan fame.
There is no evidence that Pygmalion Re-Versed is the inverse of Pygmalion; thus neither
Pygmalion Re-Versed . Pygmalion nor Pygmalion . Pygmalion Re-Versed necessarily equal I.
However, there isn’t any evidence that they aren’t inverses. But that is the subject of another
lab.
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