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 on 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 vectors--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.