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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
learning about planes

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and their equations and various
different geometric problems

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relating to them.

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So I have an example
of such a problem here.

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So I've got a
point, which happens

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to be the origin, which I'm
going to call P, (0, 0, 0).

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And I've got a plane which
has the equation 2x plus y

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minus 2z is equal to 4.

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And what I'd like you to
do is compute the distance

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from that point to that plane.

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So just to remind you, so there
are lots of points on a plane,

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of course.

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And our point in question has
a distance to each of them.

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When we talk about the distance
between a point and a plane,

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what we mean is the
shortest distance.

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So the perpendicular distance.

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So if we have the plane
and we have the point,

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so we want to drop
a perpendicular

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from the point to the
plane, and then we're

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asking for that length
of that segment.

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So that's the distance between
the point and the plane.

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So why don't you pause the
video, take a little while

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to figure this out, come
back, and we can figure it out

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together.

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So hopefully you had some
luck working out this problem.

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Let's think about
it a little bit.

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We have a point and
we have a plane,

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and we want to figure out what
the perpendicular distance

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from the point to the plane is.

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So one thing that's going
to be important then

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is definitely knowing what
direction that vector is.

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Right?

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We have the plane and we want
to find a perpendicular segment

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to it.

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And so in order to do that, it's
useful to know what direction

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is that segment pointing in.

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So luckily, we're given the
plane in this simple equation

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form.

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So the normal to
the plane-- and when

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you're given an equation
of a plane in this form,

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the normal vector is just given
by the coefficients of x, y,

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and z.

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So in our case, the plane is
2x plus y minus 2z equals 4,

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so the normal
vector to this plane

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is the vector 2, 1, minus 2.

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So this is the
direction in which

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we need to go from our point
P in order to get to the plane

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by the shortest distance.

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So now what we need
is we need to know

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the component of--
or sorry, rather, we

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need to know the
actual distance we have

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to travel in that direction.

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So one way to do this is if we
go back to our little picture

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here.

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We don't know what
this point is.

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We don't know--
when we start from P

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and head in the direction
perpendicular to the plane,

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we don't know what point we're
going to land on the plane at.

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But what we could do is,
if we knew some other point

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on the plane--
somewhere-- we could

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look at the vector connecting
P to that other point,

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and then we could project it
onto the normal direction.

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So we could take the component--

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so let's call this
other point Q.

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So if we choose any
point Q in the plane

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that we're looking at, we
could take the vector PQ

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and we can project it
onto this normal vector.

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And if we take the
component of this vector,

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project it onto
that normal vector--

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if we take the component of
this vector in the direction

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of the normal vector-- what
that will give us is exactly

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the length of this segment.

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Yeah?

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That projection will be
exactly the perpendicular

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segment we're looking for.

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And its length, the component--
or the absolute value

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of the component, perhaps--
will be exactly that distance.

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So good.

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So then we just have
to compute-- well,

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we need to find a point Q, and
we need to compute a component.

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So we need any
point on the plane.

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So, actually I'm going
to walk back over here.

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And to find a
point on the plane,

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we can just do this by
looking at the equation.

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So one way to go about
this, for example,

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is that you pick a variable
that appears in the equation.

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So x appears in the equation.

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And now you could just set all
the other variables equal to 0.

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And that will give you
something you can solve for x.

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So in particular,
you know, there's

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a point on this plane
with y equals z equals 0,

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and that point has 2x equals 4.

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So we can take, for example,
Q to be the point (2, 0, 0).

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So this is a point on the plane.

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So this is our
point on the plane,

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and so we have--
what we want to do

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is we want to project-- so
PQ, the vector from P to Q,

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we get by subtracting
the coordinates of P

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from those of Q. Q minus P. So
this is the vector [2, 0, 0].

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And we want the component
of PQ in the direction N.

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So the distance in
question is the--

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and really, when I say
component in this case,

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I mean the positive component.

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I want-- because the
distance has to be positive.

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So if I get a
negative component,

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I really want its
absolute value here.

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So the distance is the positive
component in the direction N.

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So what's that equal to?

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Well, it's just equal to the
absolute value of-- so we know

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the component of PQ in the
direction N is what we get when

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we take PQ and we dot it with
N divided by the length of N,

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and then to make sure
it's positive at the end,

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I want to throw in these
absolute values signs.

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So OK.

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So this is-- and this is now,
you know, we have our vector PQ

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and we have our
vector N, so it should

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be straightforward to compute
this final expression.

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So we know that N is
equal to 2, 1, minus 2.

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So the length of N-- which
is in the denominator here--

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is equal to the square
root of 2 squared

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plus 1 squared plus
minus 2 squared.

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And in the numerator, we have
the absolute value of PQ dot N.

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So PQ dot N is
going to be 2 times

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2, plus 0 times 1,
plus 0 times minus 2.

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OK.

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And so if we-- this is
just a fraction bar here.

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And so we simplify
that a little bit.

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So up top, we just have
4 plus 0 plus 0 is 4.

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And on the bottom, we
have the square root

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of 2 squared plus 1
squared plus 2 squared.

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That's going to be the
square root of 9, which is 3.

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So this is just equal to 4/3.

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So there we go.

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The distance in question is 4/3.

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The way we got it is we realized
that that distance is just

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the component of any
segment-- any vector--

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connecting our point
P to the plane,

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in the direction of the normal.

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So you choose any vector PQ.

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So, you know, you just have
to come up with a point

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Q on the plane, which
you can do by inspection

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from the equation.

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So that gives you a
vector that gets you

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from P to some
point on the plane,

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and then you choose
the component

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in the normal direction.

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And so once you do that, you
get this distance: your answer.

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So I'll end there.