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.