The following content is provided under a Creative
Commons license.
Your support will help MIT OpenCourseWare continue to
offer high-quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu.
PROFESSOR: Hi.
In the last unit, we introduced the notion of the
structure of vectors in terms of arrows in the plane.
And today, we want to talk about the added geometric
property of what I could call three-dimensional vectors or
three-dimensional arrows.
Now before I say that--
I guess I've already said it-- but before I say any more
about it, let me point out, of course, that this is a matter
of semantics.
Obviously an arrow, being a line, has only one dimension,
be it in the direction of the x-axis, be it in the x-y
plane, or be it in three-dimensional space.
When we say "three-dimensional vectors," we do not negate the
fact that we're talking about arrows, but rather that we are
dealing with a coordinate system which takes into
consideration the fact that the line is drawn through
three-dimensional space.
In other words, where as I call the lecture
three-dimensional vectors or arrows-- and by the way, both
in the notes and as I'm lecturing, I will very often,
whenever I write the word "vector," put "arrow" in
parentheses, whenever I write the word "arrow," put "vector"
in parentheses so that you can see the juxtaposition between
these two, the geometry versus the physical or
mathematical concept.
But the idea is something like this.
In three-dimensional space, we have a natural extension of
Cartesian coordinates.
Rather than talk about the x-y plane, we pick a third axis, a
third number line, which goes through the origin,
perpendicular to the x-y plane, and which has the sense
that if the x-axis is rotated into the y-axis through the
positive 90-degree angle here, that the z-axis is in the
direction in which a right-handed screw would turn.
I won't belabor this.
It's done very, very nicely in the textbook.
And I'm sure that you have seen this type of coordinate
system before.
If you haven't, it's something that takes a matter of some 20
or 30 seconds to pick up.
But at any rate, what we're saying is let's imagine this
three-dimensional coordinate system, three-dimensional
Cartesian coordinates.
The convention is that just as in the plane, we label the
point by its x and y components.
In three space, a point is labeled by its x, y, and z
components.
So for example, if I take a vector in three space--
meaning what?
It's a line that goes through three dimensional space.
I again shift it parallel to itself, so it
begins at my origin.
And at the risk of causing some confusion here, I still
think it's worth doing.
Let me call the vector A, and let me call the point at which
the vector terminates the point A. And let's suppose at
the point a has coordinates a1, a2, a3.
The reason that this doesn't make much difference if you
get confused is that if the vector, which originates at
the origin, terminates at the point whose coordinates a1,
a2, and a3, then the components of that
vector will have what?
The i component will be a1.
The j component will be a2.
And by the way, we let k be the unit vector in the
positive z direction.
And the k component will be a3.
To see this thing pictorially, notice the three-dimensional
effect of the vector A. It means that if we drop a
perpendicular from the head of a to the x-y plane, then you
see this distance is a1.
This distance is a2.
And this height is a3.
And notice, again, in terms of adding vectors head to tail et
cetera, notice that as a vector, this would be the
vector a1 i.
This would be the vector a2 j.
And this would be the vector a3 k.
In other words, the vector a is the sum of the three
vectors a1 i, a2 j, and a3 k.
In other words, to summarize this, the vector A is simply
what? a1 i, plus a2 j, plus a3 k.
And again notice the juxtapositioning, if you wish,
between the components of the vector and the
coordinates of the point.
Also, we can get ahold of the magnitude of this vector.
Remember, the magnitude is the length.
And how can we figure out the length of this vector?
Notice that, by the way, the geometry gets much tougher in
three-dimensional space.
But I want to show you something
interesting in a minute.
Let's suffer through the geometry for a minute.
Let's see how we can find the length of the vector A. Well,
notice, by the way, that A happens to be--
or the magnitude of A-- happens to be the hypotenuse
of a right triangle.
What right triangle is it?
It's the right triangle that joins the origin--
let me call this point here B--
and A.
In other words, triangle OBA is a right triangle because OB
is in the x-y plane and AB is
perpendicular to the x-y plane.
Therefore, by the Pythagorean theorem, the magnitude of a
will be the square root of the square of the magnitude of OB
plus the square of the magnitude of AB.
On the other hand, notice that OCB is also a right triangle.
So by the Pythagorean theorem, the length of OB is the square
root of a1 squared plus a2 squared.
And therefore, you see, by the Pythagorean theorem, the
square of the magnitude of A is this
squared plus this squared.
In other words, the magnitude of A is the square root of a1
squared, plus a2 squared, plus a3 squared.
In other words, in Cartesian coordinates, to find the
magnitude of a vector, you need only take the positive
square root of the sum of the squares of the components.
It's a regular distance formula,
the Pythagorean theorem.
By the way, just a very, very quick review from part one of
our course--
remember that technically speaking, the square root is a
double-valued function.
Namely, any positive number has two square roots.
All right?
In other words, the square root of 16 is 4.
It's also negative 4.
But by convention, when we don't indicate a sign, we're
always referring to the positive square root.
And that's fine.
Because after all, the magnitude--
the length of an arrow--
is always looked upon as being what?
Some non-negative number.
Notice that the negativeness is taken care of by the sense.
OK?
And this is what's rather interesting.
I think it's worth noting.
If we use Cartesian coordinates in one dimension,
you have is the x-axis, in which case any vector is some
scale or multiple of the unit vector i.
In other words, A is equal to a1 i.
In this case, the magnitude of A is just what?
It's the square root of a1 squared.
And notice that means the positive square root--
notice, by the way, that that's the usual meaning of
absolute value.
After all, if a1 were already positive, taking the positive
square root of its square root of its square.
These tongue twisters.
The positive square root of the square would give you the
number back again.
On the other hand, if a1 were negative, and then you square
it, and then take the positive square root, all you do is
change the sign of a1.
And since it was negative, changing its
sign makes it positive.
So this is the usual absolute value, all right?
Now in terms of the previous unit, we saw that in
two-dimensional space, using i and j as our basic vectors,
that if A was the vector a1 i plus a2 j, then the magnitude
of A, again, by the Pythagorean theorem, was just
a square root of a1 squared plus a2 squared.
And now we've seen that structurally--
even though geometrically it was harder to show--
that if A in three-dimensional space is a1 i plus a2 j plus
a3 k, then A is what?
The positive square root of a1 squared, plus a2 squared, plus
a3 squared.
And notice you see this structural resemblance.
This is a beautiful structural resemblance.
Why is it a beautiful structural resemblance?
Because now maybe we're getting the feeling that once
our rules and recipes are made up, we don't have to worry
about how difficult the geometry is.
In other words, except for the fact that you have an extra
component to worry about here, there is no basic difference
structurally between finding the length of a vector in
one-dimensional space or in three-dimensional space.
And in fact, another reason that I'm harping on this is
the fact that in a little while--
meaning within the next few lessons--
we are going to be talking about some horrible thing
called n-dimensional space, which geometrically cannot be
drawn, but which analytically preserves the same structure
that we're talking about here.
You see, the whole bag is still this idea of structure.
We're trying to emphasize that the recipes remain the same
independently of the dimension.
And just to carry this idea a few steps further, what we're
saying is, remember in the last unit, we talked about the
fact that the nice thing about i and j components was that
the definition for adding two vectors was that if they were
in terms of i and j components, you just add it
component by component?
Well again, without belaboring the point and leaving the
details to the text and to the exercises, let's keep in mind
that it's relatively easy to show that the same formal
definition of addition-- namely, placing the arrows
head to tail, et cetera--
forgetting about Cartesian coordinates, but notice the
definition of addition doesn't mention the coordinate system.
It just says put the vectors head to tail, et cetera.
And what I'm saying is it's easy to show, under those
conditions, that if A is the vector a1 i plus a2 j plus a3
k, and B is the vector b1 i plus b2 j plus b3 k, then you
still add vectors the same way in three space as
you did in two space.
Again, the geometric arguments are a little bit more
sophisticated, if only because it's harder to draw three
dimensions to scale than it is two dimensions.
But the idea is what?
A plus B is just obtained by adding what?
The two i components, the two j components, the two k
components.
And similarly, for scalar multiplication, in the same
way in terms of i and j components, you simply
multiplied each complement by the scale of c.
If c is any number and A is still the vector a1 i plus a2
j plus a3 k, the scalar multiple c times a turns out
to be c a1 i plus c a3 j plus c a3 k.
That just multiply component by component.
And by the way, don't lose track of this important point,
that the scalar multiple cA means the same thing in two
dimensions as it did three.
Because after all, when you draw that vector, once that
arrow is drawn, it's a one-dimensional thing if you
pick as your axis the line of action of
the particular arrow.
In other words, scalar multiplication
means the same thing.
What's happening is that you're looking at it in terms
of different components.
You're looking at it in terms of i, j, and k, rather than
just i and j, or in terms of any other coordinate system.
But carrying this out in particular, notice then, the
connection between the vector minus B-- negative B--
and the vector -1 times B. Namely, the vector negative B,
which has the same magnitude and direction as B, but the
opposite sense, is the same way as saying what?
Multiply B by minus 1.
Because that gives you a vector which has the same
magnitude--
namely 1 times as much as B--
the same direction, and the opposite sense.
In other words, the vector negative B, in
three-dimensional space, is minus b1 i, minus
b2 j, minus b3 k.
And therefore, since A minus B still means A plus negative B,
the vector A minus B is simply obtained how?
You subtract the same way as we did in ordinary
one-dimensional analytic geometry, in
terms of a number line.
In other words, you subtract component by component--
subtracting what?
The first vector minus the corresponding component of the
second vector.
In other words, in Cartesian coordinates, the vector A
minus B-- once we know a1, a2, and a3 as the components of A,
b1, b2, and b3 as the components of B, we just
subtract component by component to get this
particular result.
And this is, I think, very important to understand.
By the same token, as simple as this is, it's going to
cause-- if we're not careful--
great misinterpretation.
Namely, the vector arithmetic that we're talking about does
not depend on Cartesian coordinates.
Vectors or arrows, as we're looking at them, and the
arrows exist no matter what coordinate system we're using.
What is particularly important, however, is that
many of the luxuries of using vector arithmetic hinge on
using Cartesian coordinates.
Now what do I mean by that?
Well, let me just emphasize: Note the need for Cartesian
coordinates.
Let me talk instead, you see, about something called polar
coordinates.
Now later on, we're going to talk about polar coordinates
in more detail--
in fact, in very, very much more detail.
But for the time being, let's view polar coordinates simply
as a radar-type thing, as a range and bearing-type
navigation.
Namely, in polar coordinates, the way you specify a vector
in the plane, say, is you specify its length, which
we'll call r, and the angle that it makes with the
positive x-axis that we'll call theta.
In other words, in polar coordinates, the vector A,
which I've labeled r2 comma theta2 , that would indicate
that the magnitude of A is r2.
And that the angle that A makes with the
positive x-axis is theta2.
And when I label the vector B as r1 comma theta1, it's
simply my way of saying that the magnitude of the vector B
is some length, which we'll call r1.
And the angle that B makes with the positive x-axis is
some angle which we'll call theta1.
OK?
Perfectly well-defined system, isn't it?
In other words, if I tell you the range and bearing, I've
certainly given you as much information as if I told you
the i and j components.
Now here's the important point, the real
kicker to this thing.
And that is that the vector A minus B--
and A minus B is what?
It's this vector here.
And by the way, notice how vector arithmetic goes.
How could you tell quickly whether this is A minus
B or B minus A?
We talked about this in the previous unit, and had
exercises and discussions on it.
But notice again, the structure of vector
arithmetic.
If you look at addition, what is this vector here?
This is the vector which must be added on to B to give the
vector A. And that, by definition, is called what?
A minus B. A minus B is what vector?
The vector you must add on to B to yield A. Subtraction is
still the inverse of addition.
And hopefully by now, you should be becoming much less
fearful of this vector notation, that structurally,
it's the same as our first unit of arithmetic.
Well, numerical arithmetic, OK, the more regular type,
ordinary type of arithmetic.
But here's the important point.
This vector is still A minus B. But I claim that it's
trivial to see that this vector is not the vector whose
magnitude is r2 minus r1 and whose angle of bearing is
theta2 minus theta1.
Now the word "trivial" is a very misleading word in
mathematics.
One of our famous mathematical anecdotes is the professor who
said that a proof was trivial.
The student says, "It doesn't seem trivial to me." Their
professor says, "Well, it is trivial." He didn't quite see
it in a hurry.
He says, "Wait here." He ran down to his office.
Came up three hours later with a ream of papers that he had
written on, bathed in perspiration, and a big smile.
And he said, "I was right.
It was trivial."
So just because what I think is trivial may not be what you
think is trivial, let me go through this statement in more
computational detail.
What I'm saying is, let v denote the vector that we
called A minus B before.
Notice that in terms of what the vectors A and B are, this
length is r1.
This length is r2.
And notice that since this whole angle was theta2, and
the angle from here to here with theta1, this angle in
here must be the difference between those two angles,
which is theta2 minus theta1.
Now notice that this is still a triangle.
The length of this triangle is still expressible in terms of
these sides and the included angle.
But you may recall from plane trigonometry and our refresher
of that in the first part of our course--
part one--
that to find the third side of a triangle given two sides and
the included angle, one must use the Law of Cosines.
In other words, the magnitude of v has what property?
That its square is equal to the sum of the squares of
these two sides minus twice the product of these two
lengths times the cosine of the included angle.
In other words, this is just--
I hope this shows up all right-- this
is the Law of Cosines.
And if it's difficult to read, don't bother reading it.
It's still the Law of Cosines, and hopefully you
recognize it as such.
In other words, to find the magnitude of v, it's not just
r1 or r2 minus r1.
It's the square root of (r1 squared + r2 squared - 2 * r1
* r2 * cosine( theta2 - theta1)).
Can you use polar coordinates if you want?
The answer is you bet you can.
But when you do use it, make sure that whenever you're
going to find the magnitude, that you don't just subtract
the two magnitudes.
You have to use this.
And by the way, notice I haven't even gone into this
part because it's irrelevant from the point of view that
I'm trying to emphasize now.
This mess just gives you the magnitude of a vector.
It doesn't even tell you what direction it's in.
I have to still do a heck of a lot of geometry if I want to
find out what this angle here is--
a lot of work involved.
I can use polar coordinates, but I lose some of the luxury
of Cartesian coordinates.
In fact, notice that the vector whose magnitude is r2
minus r1 and whose angle of bearing is theta2 minus
theta1, can also be computed.
Namely, how do we find theta2 minus theta1?
Well up here, we have the angle theta2.
We have the angle theta1 We computed theta2 minus theta1.
1.
I can now mark off that angle here, theta2 minus theta1.
1.
I can now take a circle of radius r1--
well, why even say a circle--
mark off the length r1 onto r2, assuming r2 is the greater
of the two lengths.
That difference will be r2 minus r1.
In other words, if I swing an arc over here,
this would r1 also.
This distance here would be r2 minus r1.
I take that distance, and mark it off down here.
And what vector is this?
This is the vector, which in polar coordinates would have
its magnitude equal to r2 minus r1, and have its angle
equal to theta2 minus theta1.
And all I want you to see is no matter how you slice it,
just from this picture alone, this vector is not the same as
this vector.
And by the way, don't make the mistake of saying, gee, the
way you've draw them, they look like they could be the
same length.
Remember that even if by coincidence these two vectors
had the same length, you must remember that vector equality
requires not just the same magnitude,
but the same direction.
And even with the same direction, the same sense.
What should be obvious is, at least based on this one
picture, that the direction of this vector certainly is not
the same as the direction of this vector.
In other words, I guess I've said this many times in part
one, and even though it comes out like phony facetiousness,
I mean it quite sincerely.
Given two vectors in polar coordinates, one certainly has
the right to invent the vector (r2 - r1, theta2 - theta1).
You certainly have the right to do that, but you don't have
the right to call that A minus B. What I meant by being
facetious is, you have the right to call it that, but
you're going to be wrong.
Because A minus B has already been defined.
And all we have now is a choice of what the answer
looks like in a particular coordinate system.
And by the way, just as a quick review, assuming that
you've studied different number bases at one time in
your careers, this has come up many times in the mathematics
curriculum under the heading of "Number-versus-Numeral," or
more generally, I call it the heading of
"Name-versus-Concept."
For example, consider the number 6.
6 is this many, and no matter how you slice it again, this
many is an even number.
It breaks up into bundles of 2 with none left over.
OK?
If you were to write 6 as a base 5 numeral, it's what?
One bundle of 5 with 1 left over.
In other words, notice that the number 6 ends in an odd
digit in base 5.
In other words, the test for evenness by looking at the
last digit depends on the number base.
In other words, in an even base, a number is even if its
units digit is even.
In an odd base, a number may or may not be even, depending
on whether its units digit is even.
In other words, the actual test that you're using--
the actual convenient computational property--
depends on the base.
But the important point is that 6 is always even,
independently of a number base.
In fact, the number 6 as six tally marks was invented long
before place value enumeration.
You see what I'm driving at?
In other words, the number concept stays the same.
The concept stays the same.
What the thing looks like in terms of a particular set of
names, that's what depends on the names that you choose.
All right?
I think that is very, very important to see.
The vector properties do not change.
The convenient computational recipes do depend on what
coordinate system you're choosing.
And that's one of the reasons--
the same as in part one of this course--
why we find it very convenient whenever possible to use
Cartesian coordinates.
OK?
And let me just summarize this some more.
Structurally, this is very important here.
Structurally, arrow arithmetic is the same for both two and
three dimensions.
For example, without belaboring the point, whether
we're dealing with two-dimensional arrows or
three-dimensional arrows, A plus B is B plus A. A plus B
plus C is equal to A plus B plus C, and let me just say,
et cetera, and not belabor this particular point that
structurally, we cannot tell the difference between
two-dimensional vectors and three-dimensional vectors.
What we can tell the difference between, I guess,
is the geometry.
The geometry may be more difficult to visualize in
three space than in two space.
But structurally, we cannot tell the difference.
And I guess this is what this thing is all about, that as
you read the textbook and do your assignment, there is
going to be a tendency on your part to
try to rely on diagrams.
And boy, you have to be an expert in descriptive geometry
to be able to take a view of something in three dimensions,
and try to get its true lengths as you look along
various axes and lines of sight.
The beauty is going to be that as we continue on with this
course, we will use geometry for motivation.
But once we have motivated things by geometry, we will
extract those rules that we can use even in cases where we
can't see the picture.
In much the same way, going back to our first lecture when
we talked about why b to the 0 equals 1, we made up our rules
to make sure that they would conform to the nice
computational recipes that were prevalent in the more
simple cases.
And we are going to play this thing to the hilt.
We're going to really exploit this and explore it in future
lectures, and in fact, throughout the
rest of this course.
But more about that next time.
And until next time, good bye.
Funding for the publication of this video was provided by the
Gabriella and Paul Rosenbaum Foundation.
Help OCW continue to provide free and open access to MIT
courses by making a donation at ocw.mit.edu/donate.