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PxCxlsl_YwY
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thing.
So, let's take the vector, OP.
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PxCxlsl_YwY
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OK, so vector OP,
of course, has components x,
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PxCxlsl_YwY
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y, z.
Now, we can think of this as
|
PxCxlsl_YwY
|
actually a dot product between
OP and a mysterious vector that
|
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|
won't remain mysterious for very
long,
|
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|
namely, the vector one,
two, three.
|
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|
OK, so, this condition is the
same as OP.A equals zero,
|
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|
right?
If I take the dot product
|
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|
OPdotA I get x times one plus y
times two plus z times three.
|
PxCxlsl_YwY
|
But now, what does it mean that
the dot product between OP and A
|
PxCxlsl_YwY
|
is zero?
Well, it means that OP and A
|
PxCxlsl_YwY
|
are perpendicular.
OK, so I have this vector, A.
|
PxCxlsl_YwY
|
I'm not going to be able to
draw it realistically.
|
PxCxlsl_YwY
|
Let's say it goes this way.
Then, a point,
|
PxCxlsl_YwY
|
P, solves this equation exactly
when the vector from O to P is
|
PxCxlsl_YwY
|
perpendicular to A.
And, I claim that defines a
|
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|
plane.
For example,
|
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|
if it helps you to see it,
take a vertical vector.
|
PxCxlsl_YwY
|
What does it mean to be
perpendicular to the vertical
|
PxCxlsl_YwY
|
vector?
It means you are horizontal.
|
PxCxlsl_YwY
|
It's the horizontal plane.
Here, it's a plane that passes
|
PxCxlsl_YwY
|
through the origin and is
perpendicular to this vector,
|
PxCxlsl_YwY
|
A.
OK, so what we get is a plane
|
PxCxlsl_YwY
|
through the origin perpendicular
to A.
|
PxCxlsl_YwY
|
And, in general,
what you should remember is
|
PxCxlsl_YwY
|
that two vectors have a dot
product equal to zero if and
|
PxCxlsl_YwY
|
only if that's equivalent to the
cosine of the angle between them
|
PxCxlsl_YwY
|
is zero.
That means the angle is 90°.
|
PxCxlsl_YwY
|
That means A and B are
perpendicular.
|
PxCxlsl_YwY
|
So, we have a very fast way of
checking whether two vectors are
|
PxCxlsl_YwY
|
perpendicular.
So, one additional application
|
PxCxlsl_YwY
|
I think we'll see actually
tomorrow is to find the
|
PxCxlsl_YwY
|
components of a vector along a
certain direction.
|
PxCxlsl_YwY
|
So, I claim we can use this
intuition I gave about dot
|
PxCxlsl_YwY
|
product telling us how much to
vectors go in the same direction
|
PxCxlsl_YwY
|
to actually give a precise
meaning to the notion of
|
PxCxlsl_YwY
|
component for vector,
not just along the x,
|
PxCxlsl_YwY
|
y, or z axis,
but along any direction in
|
PxCxlsl_YwY
|
space.
So, I think I should probably
|
PxCxlsl_YwY
|
stop here.
But, I will see you tomorrow at
|
PxCxlsl_YwY
|
2:00 here, and we'll learn more
about that and about cross
|
PxCxlsl_YwY
|
products.
|
5DUQ3-Y_gX4
|
[SQUEAKING]
|
5DUQ3-Y_gX4
|
[RUSTLING]
|
5DUQ3-Y_gX4
|
[CLICKING]
|
5DUQ3-Y_gX4
|
STEVEN G. JOHNSON: So I want
to revisit the things that Alan
|
5DUQ3-Y_gX4
|
talked about, but
just a little bit more
|
5DUQ3-Y_gX4
|
slowly and a bit more--
|
5DUQ3-Y_gX4
|
just try and lay out the rules
for you as clearly as I can.
|
5DUQ3-Y_gX4
|
And what we're
going to try and do
|
5DUQ3-Y_gX4
|
is, again, just revisit
the notion of a derivative
|
5DUQ3-Y_gX4
|
to try and write it in a
way that we can generalize
|
5DUQ3-Y_gX4
|
to other kinds of objects.
|
5DUQ3-Y_gX4
|
And so I'm going
to start with 18.01
|
5DUQ3-Y_gX4
|
and then go to
18.02 and so forth.
|
5DUQ3-Y_gX4
|
So as Alan said, the key
notion of a derivative,
|
5DUQ3-Y_gX4
|
just, I think, it's easy to get
so good at taking derivatives,
|
5DUQ3-Y_gX4
|
like knowing the rule
for the derivative
|
5DUQ3-Y_gX4
|
of sine or cosine or x squared.
|
5DUQ3-Y_gX4
|
You're so good at doing them
that you forget what they are,
|
5DUQ3-Y_gX4
|
right?
|
5DUQ3-Y_gX4
|
And so the very first thing
you learned about a derivative
|
5DUQ3-Y_gX4
|
is that it's the
slope of the tangent.
|
5DUQ3-Y_gX4
|
But what that really
is is linearization.
|
5DUQ3-Y_gX4
|
So you have some arbitrary
maybe nonlinear function f of x.
|
5DUQ3-Y_gX4
|
And you're at a point x.
|
5DUQ3-Y_gX4
|
And near that
point, you're going
|
5DUQ3-Y_gX4
|
to approximate the function
with a straight line.
|
5DUQ3-Y_gX4
|
That's the tangent.
|
5DUQ3-Y_gX4
|
So it's really the linear
approximation of f.
|
5DUQ3-Y_gX4
|
And then if you move a
little bit away from x-- so
|
5DUQ3-Y_gX4
|
let me call that
delta x, so not d.
|
5DUQ3-Y_gX4
|
Delta is going to
be a finite change.
|
5DUQ3-Y_gX4
|
d is going to be
infinitesimal pretty soon.
|
5DUQ3-Y_gX4
|
But if you move it just a finite
amount, a little finite amount
|
5DUQ3-Y_gX4
|
delta x away, of course,
the function value changes.
|
5DUQ3-Y_gX4
|
But in a linear approximation,
the new function value
|
5DUQ3-Y_gX4
|
is the red dot here.
|
5DUQ3-Y_gX4
|
So that linear
approximation is, if you're
|
5DUQ3-Y_gX4
|
taking the function f
of x at x plus delta x,
|
5DUQ3-Y_gX4
|
the new value is f of x.
|
5DUQ3-Y_gX4
|
And then this
linear thing is just
|
5DUQ3-Y_gX4
|
the slope, which we call f
prime of x times delta x.
|
5DUQ3-Y_gX4
|
That's just the
definition of the slope.
|
5DUQ3-Y_gX4
|
It's the little change in
y for a little change in x.
|
5DUQ3-Y_gX4
|
And, of course, these
two terms are not exact.
|
5DUQ3-Y_gX4
|
This red dot doesn't
exactly match where
|
5DUQ3-Y_gX4
|
you are in the real function.
|
5DUQ3-Y_gX4
|
So there are also corrections.
|
5DUQ3-Y_gX4
|
But the corrections
are higher order.
|
5DUQ3-Y_gX4
|
They're terms look like delta
x squared, delta x cubed maybe.
|
5DUQ3-Y_gX4
|
Maybe if the function
is not higher--
|
5DUQ3-Y_gX4
|
it doesn't have
higher derivatives,
|
5DUQ3-Y_gX4
|
it might have square root of
delta x, so delta x to the 1.1.
|
5DUQ3-Y_gX4
|
But these are all
terms that are going
|
5DUQ3-Y_gX4
|
to be higher powers
of delta x terms
|
5DUQ3-Y_gX4
|
that, if delta x is
sufficiently small,
|
5DUQ3-Y_gX4
|
these terms will become
more and more negligible
|
5DUQ3-Y_gX4
|
compared to this linear term.
|
5DUQ3-Y_gX4
|
And a nice notation
for this that's
|
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