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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 and

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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 to be the origin--

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which I'm going to call P--

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0, 0, 0.

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

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2z is equal to 4.

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

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point to that plane.

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

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plane, 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

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plane, 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, so we want to

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drop a perpendicular from the
point to the plane, and then

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we're 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 to

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figure this out, come
back, and we can

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figure it out 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, and we want to figure

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out what the perpendicular
distance from the point

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to the plane is.

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So one thing that's going
to be important then is

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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

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segment to it.

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

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direction is that segment
pointing in.

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So luckily, we're given
the plane in this

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simple equation form.

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So the normal to the plane--

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and when you're given
an equation of a

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plane in this form--

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the normal vector is just given
by the coefficients of

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x, y, 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 is the

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vector 2, 1, minus 2.

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So this is the direction in
which we need to go from our

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point P in order to get
to the plane by

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the shortest distance.

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So now what we need is we need
to know the component of-- or

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sorry, rather-- we need to know
the actual distance we

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have to travel in
that direction.

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

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little picture 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 and head in the

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direction perpendicular to the
plane, we don't know what

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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

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point on the plane--

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somewhere--

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we could look at the vector
connecting P to that other

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point, and then we
could project it

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onto the normal direction.

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So we could take the component,
so let's call this

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other point Q. So if we choose
any point Q in the plane that

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we're looking at, we could take
the vector PQ and we can

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project it onto this
normal vector.

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And if we take the component
of this vector, project it

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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--

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what that will give
us is exactly the

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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--

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or the absolute value of the
component, perhaps--

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will be exactly that distance.

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So good.

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So then we just have to compute,
well we need to find

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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, we can just do this by

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looking at the equation.

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So one way to go about this, for
example, is that you pick

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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

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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 a point on this plane

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with y equals z equals 0, 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, and so we have, what we

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want to do is we want
to project--

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so PQ, the vector from P to PQ,
we get by subtracting the

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coordinates of P from those of
Q. Q minus P. So this is the

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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, I mean

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the positive component I want.

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Because the distance
has to be positive.

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So if I get a negative
component, I really want its

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absolute value here.

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So the distance is the positive
component in the

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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

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when we take--

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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, I want to

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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

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PQ and we have our vector
N, so it should be

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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 plus 1 squared plus

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minus 2 squared.

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

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N. So PQ dot N is going to be
2 times 2, plus 0 times 1,

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plus 0 times minus 2.

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OK.

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And so if we-- and 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--

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plus 0 plus 0 is 4.

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And on the bottom, we have the
square root of 2 squared plus

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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

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just the component
of any segment--

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any vector--

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connecting our point P
to the plane in the

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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 Q on the

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plane, which you can do by
inspection from the equation.

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So that gives you a vector that
gets you from P to some

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point on the plane, and then
you choose the component in

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the normal direction.

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And so once you do that,
you get this

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distance: your answer.

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So I'll end there.

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