JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've
begun learning about various different
ways to describe planes in three-dimensional space. In particular, there
are equations and ways you can translate between
other characterizations. So I have here four
different planes for you, described in four
different ways, and what I'd like you to
do is try and figure out what the equations for each
of these four planes are. So let me see what they are. So we've got the
first one-- part a-- we have a plane where I'm
giving you its normal vector N, which is the
vector [1, 2, 3]. And I'm going to tell
you that the plane is passing through the
point 1, 0, minus 1. In part b, I'm telling
you that the plane passes through the origin, and
also that it's parallel to two vectors. It's parallel to the vector 1,
0, minus 1, and to the vector minus 1, 2, 0. In part c, I'm telling
you that a plane passes through the points (1,
2, 0), (3, 1, 1), and (2, 0, 0). And in part d, I'm telling
you that the plane is parallel to the plane in part a,
and also that it passes through the point (1, 2, 3). So what I'd like you to do is
try, for each of these four descriptions, figure
out what the equation of the associated plane is. So why don't you pause the
video, take a few minutes, work those all out, come
back, and we can work them out together. So hopefully you had some luck
working on these problems. Let's get started. So we may as well start
with the first one. So in part a, we're given
that the normal vector N is the vector i plus
2j plus 3k, or [1, 2, 3], and that it passes through
the point P-- which I'm going to call P-- 1, 0, minus 1. So this is a form you
learned in lecture. And so it's pretty
straightforward to write down the equation here. The thing to remember is
that if a point (x, y, z) is on the plane, then we have
to have that the vector N-- the normal-- is
orthogonal to the vector connecting the point (x,
y, z) to the point we know. So that's the vector
x minus 1, y minus 0-- which is just y-- z plus 1. So N and this vector
that lies in the plane have to be orthogonal, so
their dot product has to be 0. And now you just
multiply this out. So in our case--
so N is [1, 2, 3], and you take the dot product
with x minus 1, y, z, and you get 1 times x minus
1, plus 2 times y, plus 3 times z plus 1, equals 0. So that's the
equation of the plane. You could also rewrite this
a bunch of different ways. For example, you could
multiply through and collect all the constants together. So you could write this as
x plus 2y plus 3z-- and then we've got a minus 1 plus 3,
so that's plus 2-- equals 0. So these are two
different possible forms for that equation. And you can-- you
know, sometimes people write the constant
over on this side instead of leaving 0 over there. All right. So several different,
equivalent ways to rewrite it. All right. So there's the
equation for part a. Now let's take a look at part b. So for part b we have--
let's go just back and remind ourselves what
the question was-- so we have a plane that
passes through the origin. So we know a point
on the plane, and we know that it's parallel to
the two vectors 1, 0, minus 1, and minus 1, 2, 0. OK. So we've got a point and we
have two direction vectors. And so that definitely
describes a plane for us, as long as the two
directions aren't parallel, which they aren't in this case. So the question is how
do we figure out what the equation for that plane is? Well, we have this nice
way of figuring out equations for planes when we
know a point and a normal. And we know a point,
so what would be great is if we could come up with a
normal direction to this plane. So OK. So we have two
vectors in the plane, and we want to find
a vector that's perpendicular to the plane. Well, we have a nice
tool when you're given two vectors to figure out
a vector perpendicular to both of them, and that's to
take the cross product. So our normal vector should be
the cross product of these two vectors, or, you know,
any multiple of it would do as well. So for part b, the normal N
should be the cross product of the two vectors
that are in the plane, so it should be the cross
product of 1, 0, minus 1, and-- what's the other
one-- minus 1, 2, 0. So, all right, so we
just have to compute what that cross product is. So this is a determinant
whose first row is i, j, k, and whose
second and third rows are the two vectors we're crossing. And OK, so we can
expand this out. So, if you like, so this
is i-- the coordinate of i is going to be 0 minus
minus 2, so that's 2. The coordinate of
j is going to be the negative of the determinant
of this minor, which is 0 minus minus 1 times minus 1. So the determinant of
the minor is minus 1, so the coordinate of j
is going to be plus 1. And the coordinate of
k is the determinant of this minor, which is just 2. So the normal vector in this
case is the vector [2, 1, 2]. So OK. So now we've got a normal
vector and we have a point. We were given that the plane
passes through the origin. So the equation--
using the same idea as in the previous question. So the origin is just [0, 0, 0]. That's a nice point to
know it passes through. So the equation is just 2x
plus y plus 2z is equal to 0. So this is the
equation in part b. All right. So part c, we're
given that the plane passes through three points. So once again, three points. So we have a point
in particular. We have three of them. And so what we need then
to get to the equation is we need a normal. And we saw in part
b that we could get a normal if we knew two
vectors that lay in the plane. So in this case, we
have three points. So what we'd like is to find two
vectors that lie in the plane, and then use those two vectors
to come up with a normal to the plane. So in our case that's
particularly-- well, in any case, that's
not that hard. You have three points, right? So you have three points
somewhere, P, Q, and R. And so if you want to know
two vectors in the same plane as these three points,
well, you could just take the vectors that
connect one of the points to two of others, for example. So in our case, the
plane-- since the plane passes through the points
(1, 2, 0), and (3, 1, 1), and-- what's the
last one-- (2, 0, 0). So the plane is
parallel to-- well, it doesn't matter
which one we choose, so for example, we can
say the vector that goes from here to here, so we
take this and subtract that from it, so that would give us,
for example-- 2, minus 1, 1. And we could say the vector from
here to here, so we take this and subtract that from it. And that will give
us 1, minus 2, 0. So from three points we
could get two vectors that are parallel to the plane. And we have a choice
of a point to use. We could use, for example,
the same point, (1, 2, 0), as our base point. And so then we
can go back and do exactly what we did in part b. So with those two vectors, you
can take their cross product, and find a normal
vector to the plane. So I'm not going
to do that for you. I'll leave that for
you as an exercise. Finally, in part
d, we have a plane that's parallel to
the plane in part a, and passes through (1, 2, 3). The point (1, 2, 3). So let me just rewrite
over here, parallel to-- so the plane in
part a had equation x plus 2y plus 3z
plus 2 equals 0, and passing through
the point (1, 2, 3). All right. So this is the
information that we know about our plane in this case. We know that it's parallel to
the plane with this equation, and that it passes through
the point (1, 2, 3). Well, it's parallel
to this plane. Two planes are parallel
exactly when they have the same normal vector. So remember that the
normal vector here is always going to be encoded by
these coefficients of x, y, z. In part a, this plane had
normal vector [1, 2, 3], and that [1, 2, 3] shows
up in the coefficient of x, coefficient of y,
and coefficient of z, which are 1, 2, and 3, respectively. So to get parallel planes when
you have the equation already, one thing you could do
is you can just say, oh, so that just means I leave
these coefficients the same, and I have to
change the constant. Another thing you could do
is you could just go back to our definition and say,
OK, so we know that the normal vector is [1, 2, 3]--
the vector [1, 2, 3]-- and that it passes through
the point (1, 2, 3). Either of these two
methods will work. So let me describe,
let me show you what this second, this
new method I mentioned is. So we know that the
equation of the plane has to be x plus 2y
plus 3z plus something-- that's a big question
mark there-- equal to 0. We know that the
equation of the plane is going to have
to look like this. Because it has the
same normal vector, it has to be parallel
to this plane. And so then we just
need to figure out what goes into this
box in order to make this the equation
of the right plane. Well, what else do we know? We know that it passes
through the point (1, 2, 3). So when we put in 1 for
x, 2 for y, and 3 for z, this equation has to be true. This point (1, 2, 3) has to be
a solution to this equation. So when we put in
1, 2, and 3, we have to have that 1, plus
2 times 2, plus 3 times 3, plus that same question
mark, is equal to 0. Well, this part is 1
plus 4 plus 9 is 14, so 14 plus whatever goes in
here has to be equal to 0, so this better be
equal to negative 14. Negative 14. So the equation for
the plane in that case is exactly x plus 2y plus
3z minus 14 equals 0. Now, if you didn't like that
method, the other thing you can do-- which I said
before, let me just repeat it-- is that since
it's parallel to this plane, it has the same normal vector. And we knew that the
normal vector to this plane was [1, 2, 3]. So you have a normal vector--
the vector [1, 2, 3]-- and you have a point--
the point (1, 2, 3)-- and so you can just use the
usual process given a point and a normal vector. So just to recap, we had four
different characterizations of a plane. We had a plane given in terms
of its normal vector and a point that it contains. We had a plane given
in terms of a point and two vectors parallel to it. We had a plane given in
terms of three points on it. And we had a plane given
in terms of a point and of another plane
parallel to it. So we have-- in all
these different cases, we can apply different
methods to compute the equation of our plane. So in the first case, we just
do this very straightforward computation that you
saw in lecture here. Where you just realize
that the normal vector has to be orthogonal to the
vector lying in the plane. So you take their dot
product and that gives you the equation right away. In the second
case, where you had two parallel vector-- or
sorry, yeah, two parallel vectors to the
plane, two vectors lying in the plane-- you need
to come up with a normal. And you can always come up
with a normal by taking a cross product of those two vectors,
as long as you're careful. If you accidentally chose
your two vectors parallel to each other,
that wouldn't work. You'd just get 0 here,
and that's no good. But, so you have to choose
two non-parallel vectors in the plane in order
to make this work. In the third case,
you have three points. And so with three points
what you can do is you can choose two
vectors connecting some of those points. And that gives you two
vectors that lie in the plane, and that reduces to the
case of the previous part, and then again, you can
take a cross product to get a normal vector. Finally, we did
this fourth problem where we were given a
plane parallel to it. And so you can read
off the normal vector from the coefficients of x,
y, and z in the equation. And then either use
the very first method with a point and
normal vector, or just realize that you just have
to find the appropriate value of the constant so that
this point actually lies on the plane. So I'll end there.
