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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've begun
learning about various

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different ways to describe
planes in

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three dimensional space.

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In particular, there are
equations and ways you can

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translate between other
characterizations.

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So I have here four different
planes for you described in

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four different ways, and what
I'd like you to do is try and

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figure out what the equations
for each of

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these four planes are.

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So let me see what they are.

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So we've got the first
one-- part a--

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we have a plane where I'm giving
you its normal vector

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N, which is the vector
1, 2, 3.

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And I'm going to tell you that
the plane is passing through

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the point 1, 0, minus 1.

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In part b, I'm telling you that
the plane passes through

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the origin, and also that it's
parallel to two vectors.

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It's parallel to the vector 1,
0, minus 1, and to the vector

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minus 1, 2, 0.

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In part c, I'm telling you that
a plane passes through

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the points 1, 2, 0--

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3, 1, 1--

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and 2, 0, 0.

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And in part d, I'm telling you
that the plane is parallel to

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the plane in part a, and also
that it passes through the

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point 1, 2, 3.

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So what I'd like you to do is
try for each of these four

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descriptions, figure out what
the equation of the

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associated plane is.

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So why don't you pause the
video, take a few minutes,

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work those all out, come
back, and we can

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work them out together.

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So hopefully you had some luck
working on these problems.

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Let's get started.

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So we may as well start
with the first one.

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So in part a, we're given that
the normal vector N is the

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vector i plus 2j plus 3k, or
1, 2, 3 and that it passes

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through the point P--

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which I'm going to call P--

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1, 0, minus 1.

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So this is a form you
learned in lecture.

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And so it's pretty
straightforward to write down

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the equation here.

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The thing to remember is that
if a point x, y, z is on the

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plane, then we have to have
that the vector N--

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the normal--

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is orthogonal to the vector
connecting the point x, y, z

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to the point we know.

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So that's the vector x
minus 1, y minus 0--

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which is just y--

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z plus 1.

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So N and this vector that lies
in the plane have to be

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orthogonal, so their dot
product has to be 0.

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And now you just multiply
this out.

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So in our case, so N is 1, 2,
3, and you take the dot

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product with x minus 1, y, z,
and you get 1 times x minus 1,

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plus 2 times y, plus 3 times
z plus 1, equals 0.

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So that's the equation
of the plane.

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You could also rewrite this
a bunch of different ways.

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For example, you could multiply
through and collect

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all the constants together.

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So you could write this
as x plus 2y plus 3z--

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and then we've got a minus 1
plus 3, so that's plus 2--

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equals 0.

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So these are two different
possible

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forms for that equation.

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And you can, you know, sometimes
people write the

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constant over on this
side instead of

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leaving 0 over there.

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All right.

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So several different, equivalent
ways to rewrite it.

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All right.

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So there's the equation
for part a.

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Now let's take a
look at part b.

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So for part b we have--

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let's go just back and
remind ourselves what

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the question was--

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so we have a plane that passes
through the origin.

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So we know a point on the plane,
and we know that it's

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parallel to the two vectors 1,
0, minus 1, and minus 1, 2, 0.

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OK.

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So we've got a point and we have
two direction vectors.

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And so that definitely describes
a plane for us, as

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long as the two directions
aren't parallel, which they

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aren't in this case.

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So the question is how do we
figure out what the equation

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for that plane is?

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Well, we have this nice way of
figuring out equations for

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planes when we know a
point and a normal.

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And we know a point, so what
would be great is if we could

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come up with a normal direction
to this plane.

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So OK.

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So we have two vectors in the
plane, and we want to find a

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vector that's perpendicular
to the plane.

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Well, we have a nice tool when
you're given two vectors to

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figure out a vector
perpendicular to both of them,

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and that's to take the
cross product.

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So our normal vector should be
the cross product of these two

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vectors, or, you know,
any multiple of

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it would do as well.

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So for part b, the normal N
should be the cross product of

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the two vectors that are in the
plane, so it should be the

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cross product of 1,
0, minus 1, and--

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what's the other one--
minus 1, 2, 0.

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So, all right, so we just have
to compute what that cross

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product is.

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So this is a determinant whose
first row is i, j, k, and

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whose second and third
rows are the two

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vectors we're crossing.

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And OK, so we can
expand this out.

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So, if you like,
so this is i--

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the coordinate of i is going
to be 0 minus minus

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2, so that's 2.

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The coordinate of j is going
to be the negative of the

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determinant of this minor,
which is 0 minus minus

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1 times minus 1.

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So the determinant of the
minor is minus 1, so the

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coordinate of j is going
to be plus 1.

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And the coordinate of k is the
determinant of this minor,

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which is just 2.

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So the normal vector in this
case is the vector 2, 1, 2.

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So OK.

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So now we've got a normal vector
and we have a point.

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We were given that the plane
passes through the origin.

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So the equation, using
the same idea as in

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the previous question.

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So the origin is just 0, 0, 0.

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That's a nice point to know
it passes through.

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So the equation is just 2x plus
y plus 2z is equal to 0.

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So this is the equation
in part b.

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All right.

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So part c, we're given
that the plane passes

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through three points.

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So once again, three points.

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So we have a point
in particular.

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We have three of them.

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And so what we need then
to get to the equation

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is we need a normal.

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And we saw in part b that we
could get a normal if we knew

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two vectors that lay
in the plane.

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So in this case, we
have three points.

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So what we'd like is to find
two vectors that lie in the

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plane, and then use those two
vectors to come up with a

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normal to the plane.

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So in our case that's
particularly--

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well, in any case--

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that's not that hard.

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You have three points, right?

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So you have three points
somewhere, P, Q, and R. And so

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if you want to know two vectors
in the same plane as

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these three points, well, you
could just take the vectors

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that connect one of the points
to two of others, for example.

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So in our case, the plane--

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since the plane passes through
the points 1, 2, 0, and 3, 1,

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1, and what's the last
one, 2, 0, 0.

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So the plane is parallel to--
well, it doesn't matter which

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one we choose, so for example,
we can say the vector that

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goes from here to here, so we
take this and subtract that

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from it, so that would give
us, for example--

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2, minus 1, 1.

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And we could say the vector from
here to here, so we take

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this and subtract
that from it.

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And that will give
us 1, minus 2, 0.

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So from three points we could
get two vectors that are

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parallel to the plane.

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And we have a choice
of a point to use.

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We could use, for example,
the same point 1, 2,

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0 as our base point.

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And so then we can go back
and do exactly what

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we did in part b.

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So with those two vectors, you
can take their cross product,

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and find a normal vector
to the plane.

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So I'm not going to
do that for you.

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I'll leave that for you
as an exercise.

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Finally, in part d, we have a
plane that's parallel to the

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plane in part a, and passes
through 1, 2, 3.

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The point 1, 2, 3.

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So let me just rewrite over
here, parallel to-- so the

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plane in part a had equation--

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x plus 2y plus 3z plus 2 equals
0, and passing through

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the point 1, 2, 3.

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All right.

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So this is the information that
we know about our plane

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in this case.

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We know that it's parallel to
the plane with this equation,

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and that it passes through
the point 1, 2, 3.

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Well, it's parallel
to this plane.

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Two planes are parallel exactly
when they have the

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same normal vector.

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So remember that the normal
vector here is always going to

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be encoded by these coefficients
of x, y, z.

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In part a, this plane had normal
vector 1, 2, 3, and

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that 1, 2, 3 shows up in the
coefficient of x, coefficient

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of y, and coefficient of z,
which are 1, 2, and 3,

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respectively.

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So to get parallel planes when
you have the equation already,

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one thing you could do is you
can just say, oh, so that just

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means I leave these coefficients
the same, and I

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have to change the constant.

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Another thing you could do is
you could just go back to our

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definition and say, OK, so
we know that the normal

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vector is 1, 2, 3--

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the vector 1, 2, 3--

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and that it passes through
the point 1, 2, 3.

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Either of these two
methods will work.

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So let me describe, let me show
you what this second,

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this new method I
mentioned is.

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So we know that the equation of
the plane has to be x plus

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2y plus 3z plus something--

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that's a big question
mark there--

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equal to 0.

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We know that the equation of the
plane is going to have to

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look like this.

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Because it has the same normal
vector, it has to be parallel

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to this plane.

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And so then we just need to
figure out what goes into this

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box in order to make this the
equation of the right plane.

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Well, what else do we know?

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We know that it passes through
the point 1, 2, 3.

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So when we put in 1 for x, 2
for y, and 3 for z, this

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equation has to be true.

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This point 1, 2, 3 has to be a
solution to this equation.

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So when we put in 1, 2, and 3,
we have to have that 1, plus 2

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times 2, plus 3 times 3, plus
that same question mark, is

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equal to 0.

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Well, this part is 1 plus 4
plus 9 is 14, so 14 plus

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whatever goes in here has to be
equal to 0, so this better

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be equal to negative 14.

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Negative 14.

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So the equation for the plane in
that case is exactly x plus

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2y plus 3z minus 14 equals 0.

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Now, if you didn't like that
method, the other thing you

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can do-- which I said before,
let me just repeat it-- is

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that since it's parallel to this
plane, it has the same

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normal vector.

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And we knew that the normal
vector to this

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plane was 1, 2, 3.

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So you have a normal vector--
the vector 1, 2, 3-- and you

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have a point--

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the point 1, 2, 3-- and so you
can just use the usual process

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given a point and
a normal vector.

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So just to recap, we
had four different

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characterizations of a plane.

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We had a plane given in terms
of its normal vector and a

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point that it contains.

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We had a plane given in terms
of a point and two vectors

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parallel to it.

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00:12:13,300 --> 00:12:17,270
We had a plane given in terms
of three points on it.

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And we had a plane given in
terms of a point and of

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another plane parallel to it.

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So we have in all these
different cases, we can apply

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different methods to compute
the equation of our plane.

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So in the first case, we just
do this very straightforward

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computation that you saw
on lecture here.

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Where you just realize that the
normal vector has to be

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orthogonal to the vector
lying in the plane.

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So you take their dot product
and that gives you the

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equation right away.

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In the second case, where you
had two parallel vectors--or

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sorry, yeah--

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two parallel vectors to the
plane-- two vectors lying in

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00:12:51,270 --> 00:12:51,900
the plane--

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00:12:51,900 --> 00:12:53,430
you need to come up
with a normal.

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00:12:53,430 --> 00:12:55,370
And you can always come up
with a normal by taking a

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00:12:55,370 --> 00:12:57,260
cross product of those
two vectors, as

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00:12:57,260 --> 00:12:58,170
long as you're careful.

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If you accidentally chose your
two vectors parallel to each

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other, that wouldn't work.

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You'd just get 0 here,
and that's no good.

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But, so you have to choose two
non-parallel vectors in the

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plane in order to
make this work.

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In the third case, you
have three points.

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And so with three points what
you can do is you can choose

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two vectors connecting
some of those points.

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And that gives you two vectors
that lie in the plane, and

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that reduces to the case of the
previous part, and then

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00:13:22,280 --> 00:13:23,830
again, you can take
a cross product to

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00:13:23,830 --> 00:13:25,050
get a normal vector.

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00:13:25,050 --> 00:13:28,330
Finally, we did this fourth
problem where we were given a

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plane parallel to it.

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00:13:29,570 --> 00:13:32,890
And so you can read off the
normal vector from the

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00:13:32,890 --> 00:13:35,380
coefficients of x, y, and
z in the equation.

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00:13:35,380 --> 00:13:40,790
And then either use the very
first method with a point and

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00:13:40,790 --> 00:13:44,370
normal vector, or just realize
that you just have to find the

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00:13:44,370 --> 00:13:47,030
appropriate value of the
constant so that this point

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actually lies on the plane.

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So I'll end there.

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