JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been
learning about planes
and their equations and various
different geometric problems
relating to them.
So I have an example
of such a problem here.
So I've got a
point, which happens
to be the origin, which I'm
going to call P, (0, 0, 0).
And I've got a plane which
has the equation 2x plus y
minus 2z is equal to 4.
And what I'd like you to
do is compute the distance
from that point to that plane.
So just to remind you, so there
are lots of points on a plane,
of course.
And our point in question has
a distance to each of them.
When we talk about the distance
between a point and a plane,
what we mean is the
shortest distance.
So the perpendicular distance.
So if we have the plane
and we have the point,
so we want to drop
a perpendicular
from the point to the
plane, and then we're
asking for that length
of that segment.
So that's the distance between
the point and the plane.
So why don't you pause the
video, take a little while
to figure this out, come
back, and we can figure it out
together.
So hopefully you had some
luck working out this problem.
Let's think about
it a little bit.
We have a point and
we have a plane,
and we want to figure out what
the perpendicular distance
from the point to the plane is.
So one thing that's going
to be important then
is definitely knowing what
direction that vector is.
Right?
We have the plane and we want
to find a perpendicular segment
to it.
And so in order to do that, it's
useful to know what direction
is that segment pointing in.
So luckily, we're given the
plane in this simple equation
form.
So the normal to
the plane-- and when
you're given an equation
of a plane in this form,
the normal vector is just given
by the coefficients of x, y,
and z.
So in our case, the plane is
2x plus y minus 2z equals 4,
so the normal
vector to this plane
is the vector 2, 1, minus 2.
So this is the
direction in which
we need to go from our point
P in order to get to the plane
by the shortest distance.
So now what we need
is we need to know
the component of--
or sorry, rather, we
need to know the
actual distance we have
to travel in that direction.
So one way to do this is if we
go back to our little picture
here.
We don't know what
this point is.
We don't know--
when we start from P
and head in the direction
perpendicular to the plane,
we don't know what point we're
going to land on the plane at.
But what we could do is,
if we knew some other point
on the plane--
somewhere-- we could
look at the vector connecting
P to that other point,
and then we could project it
onto the normal direction.
So we could take the component--
so let's call this
other point Q.
So if we choose any
point Q in the plane
that we're looking at, we
could take the vector PQ
and we can project it
onto this normal vector.
And if we take the
component of this vector,
project it onto
that normal vector--
if we take the component of
this vector in the direction
of the normal vector-- what
that will give us is exactly
the length of this segment.
Yeah?
That projection will be
exactly the perpendicular
segment we're looking for.
And its length, the component--
or the absolute value
of the component, perhaps--
will be exactly that distance.
So good.
So then we just have
to compute-- well,
we need to find a point Q, and
we need to compute a component.
So we need any
point on the plane.
So, actually I'm going
to walk back over here.
And to find a
point on the plane,
we can just do this by
looking at the equation.
So one way to go about
this, for example,
is that you pick a variable
that appears in the equation.
So x appears in the equation.
And now you could just set all
the other variables equal to 0.
And that will give you
something you can solve for x.
So in particular,
you know, there's
a point on this plane
with y equals z equals 0,
and that point has 2x equals 4.
So we can take, for example,
Q to be the point (2, 0, 0).
So this is a point on the plane.
So this is our
point on the plane,
and so we have--
what we want to do
is we want to project-- so
PQ, the vector from P to Q,
we get by subtracting
the coordinates of P
from those of Q. Q minus P. So
this is the vector [2, 0, 0].
And we want the component
of PQ in the direction N.
So the distance in
question is the--
and really, when I say
component in this case,
I mean the positive component.
I want-- because the
distance has to be positive.
So if I get a
negative component,
I really want its
absolute value here.
So the distance is the positive
component in the direction N.
So what's that equal to?
Well, it's just equal to the
absolute value of-- so we know
the component of PQ in the
direction N is what we get when
we take PQ and we dot it with
N divided by the length of N,
and then to make sure
it's positive at the end,
I want to throw in these
absolute values signs.
So OK.
So this is-- and this is now,
you know, we have our vector PQ
and we have our
vector N, so it should
be straightforward to compute
this final expression.
So we know that N is
equal to 2, 1, minus 2.
So the length of N-- which
is in the denominator here--
is equal to the square
root of 2 squared
plus 1 squared plus
minus 2 squared.
And in the numerator, we have
the absolute value of PQ dot N.
So PQ dot N is
going to be 2 times
2, plus 0 times 1,
plus 0 times minus 2.
OK.
And so if we-- this is
just a fraction bar here.
And so we simplify
that a little bit.
So up top, we just have
4 plus 0 plus 0 is 4.
And on the bottom, we
have the square root
of 2 squared plus 1
squared plus 2 squared.
That's going to be the
square root of 9, which is 3.
So this is just equal to 4/3.
So there we go.
The distance in question is 4/3.
The way we got it is we realized
that that distance is just
the component of any
segment-- any vector--
connecting our point
P to the plane,
in the direction of the normal.
So you choose any vector PQ.
So, you know, you just have
to come up with a point
Q on the plane, which
you can do by inspection
from the equation.
So that gives you a
vector that gets you
from P to some
point on the plane,
and then you choose
the component
in the normal direction.
And so once you do that, you
get this distance: your answer.
So I'll end there.
