WEBVTT

00:00:06.815 --> 00:00:07.440
JOEL LEWIS: Hi.

00:00:07.440 --> 00:00:08.950
Welcome back to recitation.

00:00:08.950 --> 00:00:10.390
In lecture, you've
begun learning

00:00:10.390 --> 00:00:12.940
about various different
ways to describe planes

00:00:12.940 --> 00:00:14.380
in three-dimensional space.

00:00:14.380 --> 00:00:16.310
In particular, there
are equations and ways

00:00:16.310 --> 00:00:18.830
you can translate between
other characterizations.

00:00:18.830 --> 00:00:20.830
So I have here four
different planes for you,

00:00:20.830 --> 00:00:22.570
described in four
different ways,

00:00:22.570 --> 00:00:24.640
and what I'd like you to
do is try and figure out

00:00:24.640 --> 00:00:27.210
what the equations for each
of these four planes are.

00:00:27.210 --> 00:00:28.730
So let me see what they are.

00:00:28.730 --> 00:00:30.570
So we've got the
first one-- part

00:00:30.570 --> 00:00:33.510
a-- we have a plane where I'm
giving you its normal vector

00:00:33.510 --> 00:00:36.119
N, which is the
vector [1, 2,  3].

00:00:36.119 --> 00:00:37.910
And I'm going to tell
you that the plane is

00:00:37.910 --> 00:00:41.570
passing through the
point 1, 0, minus 1.

00:00:41.570 --> 00:00:44.820
In part b, I'm telling
you that the plane passes

00:00:44.820 --> 00:00:47.450
through the origin, and
also that it's parallel

00:00:47.450 --> 00:00:48.210
to two vectors.

00:00:48.210 --> 00:00:51.900
It's parallel to the vector 1,
0, minus 1, and to the vector

00:00:51.900 --> 00:00:54.510
minus 1, 2, 0.

00:00:54.510 --> 00:00:58.200
In part c, I'm telling
you that a plane

00:00:58.200 --> 00:01:04.370
passes through the points (1,
2, 0), (3, 1, 1), and (2, 0, 0).

00:01:04.370 --> 00:01:07.540
And in part d, I'm telling
you that the plane is parallel

00:01:07.540 --> 00:01:10.550
to the plane in part a,
and also that it passes

00:01:10.550 --> 00:01:12.770
through the point (1, 2, 3).

00:01:12.770 --> 00:01:15.430
So what I'd like you to do is
try, for each of these four

00:01:15.430 --> 00:01:17.550
descriptions, figure
out what the equation

00:01:17.550 --> 00:01:19.600
of the associated plane is.

00:01:19.600 --> 00:01:21.980
So why don't you pause the
video, take a few minutes,

00:01:21.980 --> 00:01:24.840
work those all out, come
back, and we can work them out

00:01:24.840 --> 00:01:25.340
together.

00:01:33.450 --> 00:01:35.830
So hopefully you had some luck
working on these problems.

00:01:35.830 --> 00:01:37.850
Let's get started.

00:01:37.850 --> 00:01:40.670
So we may as well start
with the first one.

00:01:40.670 --> 00:01:44.590
So in part a, we're given
that the normal vector

00:01:44.590 --> 00:01:50.480
N is the vector i plus
2j plus 3k, or [1, 2, 3],

00:01:50.480 --> 00:01:54.240
and that it passes through
the point P-- which I'm going

00:01:54.240 --> 00:01:58.370
to call P-- 1, 0, minus 1.

00:01:58.370 --> 00:02:01.180
So this is a form you
learned in lecture.

00:02:01.180 --> 00:02:02.850
And so it's pretty
straightforward

00:02:02.850 --> 00:02:04.530
to write down the equation here.

00:02:04.530 --> 00:02:06.940
The thing to remember is
that if a point (x, y, z)

00:02:06.940 --> 00:02:10.890
is on the plane, then we have
to have that the vector N--

00:02:10.890 --> 00:02:14.380
the normal-- is
orthogonal to the vector

00:02:14.380 --> 00:02:17.370
connecting the point (x,
y, z) to the point we know.

00:02:17.370 --> 00:02:23.310
So that's the vector
x minus 1, y minus 0--

00:02:23.310 --> 00:02:25.950
which is just y-- z plus 1.

00:02:25.950 --> 00:02:28.800
So N and this vector
that lies in the plane

00:02:28.800 --> 00:02:32.010
have to be orthogonal, so
their dot product has to be 0.

00:02:32.010 --> 00:02:33.640
And now you just
multiply this out.

00:02:33.640 --> 00:02:36.950
So in our case--
so N is [1, 2,  3],

00:02:36.950 --> 00:02:39.800
and you take the dot product
with x minus 1, y, z,

00:02:39.800 --> 00:02:47.300
and you get 1 times x minus
1, plus 2 times y, plus 3

00:02:47.300 --> 00:02:50.710
times z plus 1, equals 0.

00:02:50.710 --> 00:02:52.390
So that's the
equation of the plane.

00:02:52.390 --> 00:02:54.640
You could also rewrite this
a bunch of different ways.

00:02:54.640 --> 00:02:57.050
For example, you could
multiply through and collect

00:02:57.050 --> 00:02:58.340
all the constants together.

00:02:58.340 --> 00:03:04.780
So you could write this as
x plus 2y plus 3z-- and then

00:03:04.780 --> 00:03:09.640
we've got a minus 1 plus 3,
so that's plus 2-- equals 0.

00:03:09.640 --> 00:03:11.390
So these are two
different possible forms

00:03:11.390 --> 00:03:12.140
for that equation.

00:03:12.140 --> 00:03:13.515
And you can-- you
know, sometimes

00:03:13.515 --> 00:03:15.993
people write the constant
over on this side instead

00:03:15.993 --> 00:03:16.993
of leaving 0 over there.

00:03:16.993 --> 00:03:17.450
All right.

00:03:17.450 --> 00:03:19.640
So several different,
equivalent ways to rewrite it.

00:03:19.640 --> 00:03:20.430
All right.

00:03:20.430 --> 00:03:22.560
So there's the
equation for part a.

00:03:22.560 --> 00:03:24.720
Now let's take a look at part b.

00:03:24.720 --> 00:03:28.260
So for part b we have--
let's go just back

00:03:28.260 --> 00:03:30.340
and remind ourselves what
the question was-- so

00:03:30.340 --> 00:03:33.000
we have a plane that
passes through the origin.

00:03:33.000 --> 00:03:34.840
So we know a point
on the plane, and we

00:03:34.840 --> 00:03:38.900
know that it's parallel to
the two vectors 1, 0, minus 1,

00:03:38.900 --> 00:03:41.611
and minus 1, 2, 0.

00:03:41.611 --> 00:03:42.110
OK.

00:03:42.110 --> 00:03:45.940
So we've got a point and we
have two direction vectors.

00:03:45.940 --> 00:03:49.560
And so that definitely
describes a plane for us,

00:03:49.560 --> 00:03:52.170
as long as the two
directions aren't parallel,

00:03:52.170 --> 00:03:53.607
which they aren't in this case.

00:03:53.607 --> 00:03:55.440
So the question is how
do we figure out what

00:03:55.440 --> 00:03:57.380
the equation for that plane is?

00:03:57.380 --> 00:03:59.300
Well, we have this nice
way of figuring out

00:03:59.300 --> 00:04:02.800
equations for planes when we
know a point and a normal.

00:04:02.800 --> 00:04:04.880
And we know a point,
so what would be great

00:04:04.880 --> 00:04:08.181
is if we could come up with a
normal direction to this plane.

00:04:08.181 --> 00:04:08.680
So OK.

00:04:08.680 --> 00:04:12.050
So we have two
vectors in the plane,

00:04:12.050 --> 00:04:13.790
and we want to find
a vector that's

00:04:13.790 --> 00:04:15.330
perpendicular to the plane.

00:04:15.330 --> 00:04:17.010
Well, we have a nice
tool when you're

00:04:17.010 --> 00:04:20.080
given two vectors to figure out
a vector perpendicular to both

00:04:20.080 --> 00:04:22.020
of them, and that's to
take the cross product.

00:04:22.020 --> 00:04:25.590
So our normal vector should be
the cross product of these two

00:04:25.590 --> 00:04:29.350
vectors, or, you know,
any multiple of it

00:04:29.350 --> 00:04:32.190
would do as well.

00:04:32.190 --> 00:04:39.170
So for part b, the normal N
should be the cross product

00:04:39.170 --> 00:04:40.930
of the two vectors
that are in the plane,

00:04:40.930 --> 00:04:46.150
so it should be the cross
product of 1, 0, minus 1,

00:04:46.150 --> 00:04:51.370
and-- what's the other
one-- minus 1, 2, 0.

00:04:51.370 --> 00:04:53.450
So, all right, so we
just have to compute

00:04:53.450 --> 00:04:55.360
what that cross product is.

00:04:55.360 --> 00:05:07.430
So this is a determinant
whose first row

00:05:07.430 --> 00:05:10.560
is i, j, k, and whose
second and third rows are

00:05:10.560 --> 00:05:12.160
the two vectors we're crossing.

00:05:12.160 --> 00:05:14.990
And OK, so we can
expand this out.

00:05:14.990 --> 00:05:17.930
So, if you like, so this
is i-- the coordinate of i

00:05:17.930 --> 00:05:22.795
is going to be 0 minus
minus 2, so that's 2.

00:05:22.795 --> 00:05:24.710
The coordinate of
j is going to be

00:05:24.710 --> 00:05:27.530
the negative of the determinant
of this minor, which is

00:05:27.530 --> 00:05:31.015
0 minus minus 1 times minus 1.

00:05:31.015 --> 00:05:34.470
So the determinant of
the minor is minus 1,

00:05:34.470 --> 00:05:38.620
so the coordinate of j
is going to be plus 1.

00:05:38.620 --> 00:05:40.370
And the coordinate of
k is the determinant

00:05:40.370 --> 00:05:44.060
of this minor, which is just 2.

00:05:44.060 --> 00:05:48.380
So the normal vector in this
case is the vector [2, 1, 2].

00:05:48.380 --> 00:05:49.130
So OK.

00:05:49.130 --> 00:05:51.589
So now we've got a normal
vector and we have a point.

00:05:51.589 --> 00:05:53.880
We were given that the plane
passes through the origin.

00:05:53.880 --> 00:06:00.719
So the equation--
using the same idea

00:06:00.719 --> 00:06:01.885
as in the previous question.

00:06:01.885 --> 00:06:04.050
So the origin is just [0, 0, 0].

00:06:04.050 --> 00:06:06.850
That's a nice point to
know it passes through.

00:06:06.850 --> 00:06:14.420
So the equation is just 2x
plus y plus 2z is equal to 0.

00:06:14.420 --> 00:06:17.440
So this is the
equation in part b.

00:06:17.440 --> 00:06:17.940
All right.

00:06:17.940 --> 00:06:21.630
So part c, we're
given that the plane

00:06:21.630 --> 00:06:23.950
passes through three points.

00:06:23.950 --> 00:06:25.650
So once again, three points.

00:06:25.650 --> 00:06:27.240
So we have a point
in particular.

00:06:27.240 --> 00:06:29.520
We have three of them.

00:06:29.520 --> 00:06:32.430
And so what we need then
to get to the equation

00:06:32.430 --> 00:06:34.560
is we need a normal.

00:06:34.560 --> 00:06:37.440
And we saw in part
b that we could

00:06:37.440 --> 00:06:41.120
get a normal if we knew two
vectors that lay in the plane.

00:06:41.120 --> 00:06:43.250
So in this case, we
have three points.

00:06:43.250 --> 00:06:47.210
So what we'd like is to find two
vectors that lie in the plane,

00:06:47.210 --> 00:06:50.970
and then use those two vectors
to come up with a normal

00:06:50.970 --> 00:06:51.820
to the plane.

00:06:51.820 --> 00:06:55.260
So in our case that's
particularly-- well,

00:06:55.260 --> 00:06:56.970
in any case, that's
not that hard.

00:06:56.970 --> 00:06:58.650
You have three points, right?

00:06:58.650 --> 00:07:01.880
So you have three points
somewhere, P, Q, and R.

00:07:01.880 --> 00:07:05.459
And so if you want to know
two vectors in the same plane

00:07:05.459 --> 00:07:07.250
as these three points,
well, you could just

00:07:07.250 --> 00:07:11.460
take the vectors that
connect one of the points

00:07:11.460 --> 00:07:13.200
to two of others, for example.

00:07:13.200 --> 00:07:24.460
So in our case, the
plane-- since the plane

00:07:24.460 --> 00:07:34.150
passes through the points
(1, 2, 0), and (3, 1, 1),

00:07:34.150 --> 00:07:36.450
and-- what's the
last one-- (2, 0, 0).

00:07:40.320 --> 00:07:45.640
So the plane is
parallel to-- well, it

00:07:45.640 --> 00:07:48.230
doesn't matter
which one we choose,

00:07:48.230 --> 00:07:50.460
so for example, we can
say the vector that

00:07:50.460 --> 00:07:53.960
goes from here to here, so we
take this and subtract that

00:07:53.960 --> 00:07:59.890
from it, so that would give us,
for example-- 2, minus 1, 1.

00:07:59.890 --> 00:08:02.800
And we could say the vector from
here to here, so we take this

00:08:02.800 --> 00:08:05.890
and subtract that from it.

00:08:05.890 --> 00:08:12.270
And that will give
us 1, minus 2, 0.

00:08:12.270 --> 00:08:15.130
So from three points we
could get two vectors

00:08:15.130 --> 00:08:16.580
that are parallel to the plane.

00:08:16.580 --> 00:08:19.070
And we have a choice
of a point to use.

00:08:19.070 --> 00:08:21.850
We could use, for example,
the same point, (1, 2, 0),

00:08:21.850 --> 00:08:22.665
as our base point.

00:08:22.665 --> 00:08:24.040
And so then we
can go back and do

00:08:24.040 --> 00:08:25.830
exactly what we did in part b.

00:08:25.830 --> 00:08:29.650
So with those two vectors, you
can take their cross product,

00:08:29.650 --> 00:08:31.290
and find a normal
vector to the plane.

00:08:31.290 --> 00:08:32.790
So I'm not going
to do that for you.

00:08:32.790 --> 00:08:35.650
I'll leave that for
you as an exercise.

00:08:35.650 --> 00:08:41.270
Finally, in part
d, we have a plane

00:08:41.270 --> 00:08:45.330
that's parallel to
the plane in part a,

00:08:45.330 --> 00:08:47.950
and passes through (1, 2, 3).

00:08:47.950 --> 00:08:49.080
The point (1, 2, 3).

00:08:49.080 --> 00:08:53.100
So let me just rewrite
over here, parallel

00:08:53.100 --> 00:08:56.730
to-- so the plane in
part a had equation

00:08:56.730 --> 00:09:07.930
x plus 2y plus 3z
plus 2 equals 0,

00:09:07.930 --> 00:09:13.400
and passing through
the point (1, 2, 3).

00:09:13.400 --> 00:09:13.966
All right.

00:09:13.966 --> 00:09:15.590
So this is the
information that we know

00:09:15.590 --> 00:09:17.100
about our plane in this case.

00:09:17.100 --> 00:09:20.060
We know that it's parallel to
the plane with this equation,

00:09:20.060 --> 00:09:23.840
and that it passes through
the point (1, 2, 3).

00:09:23.840 --> 00:09:26.410
Well, it's parallel
to this plane.

00:09:26.410 --> 00:09:28.290
Two planes are parallel
exactly when they

00:09:28.290 --> 00:09:30.390
have the same normal vector.

00:09:30.390 --> 00:09:32.670
So remember that the
normal vector here

00:09:32.670 --> 00:09:36.070
is always going to be encoded by
these coefficients of x, y, z.

00:09:36.070 --> 00:09:41.110
In part a, this plane had
normal vector [1, 2, 3],

00:09:41.110 --> 00:09:43.260
and that [1, 2, 3] shows
up in the coefficient

00:09:43.260 --> 00:09:45.800
of x, coefficient of y,
and coefficient of z, which

00:09:45.800 --> 00:09:48.400
are 1, 2, and 3, respectively.

00:09:48.400 --> 00:09:52.820
So to get parallel planes when
you have the equation already,

00:09:52.820 --> 00:09:54.870
one thing you could do
is you can just say, oh,

00:09:54.870 --> 00:09:57.720
so that just means I leave
these coefficients the same,

00:09:57.720 --> 00:09:59.950
and I have to
change the constant.

00:09:59.950 --> 00:10:02.240
Another thing you could do
is you could just go back

00:10:02.240 --> 00:10:04.614
to our definition and say,
OK, so we know that the normal

00:10:04.614 --> 00:10:07.880
vector is [1, 2, 3]--
the vector [1, 2, 3]--

00:10:07.880 --> 00:10:11.730
and that it passes through
the point (1, 2, 3).

00:10:11.730 --> 00:10:13.730
Either of these two
methods will work.

00:10:13.730 --> 00:10:15.690
So let me describe,
let me show you

00:10:15.690 --> 00:10:18.960
what this second, this
new method I mentioned is.

00:10:18.960 --> 00:10:21.410
So we know that the
equation of the plane

00:10:21.410 --> 00:10:29.970
has to be x plus 2y
plus 3z plus something--

00:10:29.970 --> 00:10:35.710
that's a big question
mark there-- equal to 0.

00:10:35.710 --> 00:10:37.620
We know that the
equation of the plane

00:10:37.620 --> 00:10:39.079
is going to have
to look like this.

00:10:39.079 --> 00:10:40.661
Because it has the
same normal vector,

00:10:40.661 --> 00:10:42.300
it has to be parallel
to this plane.

00:10:42.300 --> 00:10:43.883
And so then we just
need to figure out

00:10:43.883 --> 00:10:46.370
what goes into this
box in order to make

00:10:46.370 --> 00:10:48.340
this the equation
of the right plane.

00:10:48.340 --> 00:10:49.680
Well, what else do we know?

00:10:49.680 --> 00:10:52.490
We know that it passes
through the point (1, 2, 3).

00:10:52.490 --> 00:10:56.650
So when we put in 1 for
x, 2 for y, and 3 for z,

00:10:56.650 --> 00:11:00.190
this equation has to be true.

00:11:00.190 --> 00:11:03.620
This point (1, 2, 3) has to be
a solution to this equation.

00:11:03.620 --> 00:11:06.030
So when we put in
1, 2, and 3, we

00:11:06.030 --> 00:11:10.750
have to have that 1, plus
2 times 2, plus 3 times

00:11:10.750 --> 00:11:16.280
3, plus that same question
mark, is equal to 0.

00:11:16.280 --> 00:11:20.520
Well, this part is 1
plus 4 plus 9 is 14,

00:11:20.520 --> 00:11:24.230
so 14 plus whatever goes in
here has to be equal to 0,

00:11:24.230 --> 00:11:26.090
so this better be
equal to negative 14.

00:11:26.090 --> 00:11:26.590
Negative 14.

00:11:30.180 --> 00:11:31.970
So the equation for
the plane in that case

00:11:31.970 --> 00:11:38.210
is exactly x plus 2y plus
3z minus 14 equals 0.

00:11:38.210 --> 00:11:41.360
Now, if you didn't like that
method, the other thing you can

00:11:41.360 --> 00:11:43.090
do-- which I said
before, let me just

00:11:43.090 --> 00:11:46.150
repeat it-- is that since
it's parallel to this plane,

00:11:46.150 --> 00:11:47.890
it has the same normal vector.

00:11:47.890 --> 00:11:50.550
And we knew that the
normal vector to this plane

00:11:50.550 --> 00:11:51.740
was [1, 2, 3].

00:11:51.740 --> 00:11:54.500
So you have a normal vector--
the vector [1, 2, 3]--

00:11:54.500 --> 00:11:56.720
and you have a point--
the point (1, 2, 3)--

00:11:56.720 --> 00:11:59.190
and so you can just use the
usual process given a point

00:11:59.190 --> 00:12:01.570
and a normal vector.

00:12:01.570 --> 00:12:04.800
So just to recap, we had four
different characterizations

00:12:04.800 --> 00:12:05.380
of a plane.

00:12:05.380 --> 00:12:08.390
We had a plane given in terms
of its normal vector and a point

00:12:08.390 --> 00:12:09.470
that it contains.

00:12:09.470 --> 00:12:11.580
We had a plane given
in terms of a point

00:12:11.580 --> 00:12:13.300
and two vectors parallel to it.

00:12:13.300 --> 00:12:17.270
We had a plane given in
terms of three points on it.

00:12:17.270 --> 00:12:19.470
And we had a plane given
in terms of a point

00:12:19.470 --> 00:12:21.570
and of another plane
parallel to it.

00:12:21.570 --> 00:12:23.960
So we have-- in all
these different cases,

00:12:23.960 --> 00:12:26.760
we can apply different
methods to compute

00:12:26.760 --> 00:12:28.150
the equation of our plane.

00:12:28.150 --> 00:12:31.530
So in the first case, we just
do this very straightforward

00:12:31.530 --> 00:12:34.010
computation that you
saw in lecture here.

00:12:34.010 --> 00:12:36.850
Where you just realize
that the normal vector has

00:12:36.850 --> 00:12:40.010
to be orthogonal to the
vector lying in the plane.

00:12:40.010 --> 00:12:42.010
So you take their dot
product and that gives you

00:12:42.010 --> 00:12:43.710
the equation right away.

00:12:43.710 --> 00:12:45.215
In the second
case, where you had

00:12:45.215 --> 00:12:49.380
two parallel vector-- or
sorry, yeah, two parallel

00:12:49.380 --> 00:12:50.840
vectors to the
plane, two vectors

00:12:50.840 --> 00:12:53.364
lying in the plane-- you need
to come up with a normal.

00:12:53.364 --> 00:12:55.780
And you can always come up
with a normal by taking a cross

00:12:55.780 --> 00:12:58.170
product of those two vectors,
as long as you're careful.

00:12:58.170 --> 00:13:00.460
If you accidentally chose
your two vectors parallel

00:13:00.460 --> 00:13:02.490
to each other,
that wouldn't work.

00:13:02.490 --> 00:13:04.610
You'd just get 0 here,
and that's no good.

00:13:04.610 --> 00:13:07.000
But, so you have to choose
two non-parallel vectors

00:13:07.000 --> 00:13:09.380
in the plane in order
to make this work.

00:13:09.380 --> 00:13:11.340
In the third case,
you have three points.

00:13:11.340 --> 00:13:13.310
And so with three points
what you can do is you

00:13:13.310 --> 00:13:15.440
can choose two
vectors connecting

00:13:15.440 --> 00:13:16.510
some of those points.

00:13:16.510 --> 00:13:19.290
And that gives you two
vectors that lie in the plane,

00:13:19.290 --> 00:13:21.830
and that reduces to the
case of the previous part,

00:13:21.830 --> 00:13:23.750
and then again, you can
take a cross product

00:13:23.750 --> 00:13:25.050
to get a normal vector.

00:13:25.050 --> 00:13:27.450
Finally, we did
this fourth problem

00:13:27.450 --> 00:13:29.570
where we were given a
plane parallel to it.

00:13:29.570 --> 00:13:32.620
And so you can read
off the normal vector

00:13:32.620 --> 00:13:35.380
from the coefficients of x,
y, and z in the equation.

00:13:35.380 --> 00:13:39.770
And then either use
the very first method

00:13:39.770 --> 00:13:42.480
with a point and
normal vector, or just

00:13:42.480 --> 00:13:45.320
realize that you just have
to find the appropriate value

00:13:45.320 --> 00:13:47.400
of the constant so that
this point actually

00:13:47.400 --> 00:13:48.550
lies on the plane.

00:13:48.550 --> 00:13:50.397
So I'll end there.