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DAVID JORDAN: Hello, and welcome
back to recitation.

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In this problem, I'd
like you to compute

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the area of a triangle.

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This triangle sits in space and
it has its three vertices,

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labeled here as P1,
P2, and P3.

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So we're going to compute this
area, and we're going to do it

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using the cross product, which
we learned about in lecture.

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So why don't you take some time
to work this out, pause

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the video, and we'll check
back in a minute and

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see how I did it.

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Hello and welcome back.

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So the first thing that I like
to do with a problem like this

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is I like to draw a picture so
I can kind of think about

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what's going on.

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So we have this triangle
sitting out in space.

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And we know that we want to take
a cross product in order

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to compute its area, but
we need to be careful.

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Cross product, it doesn't
make sense to take

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cross product of points.

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What makes sense is to take
cross product of vectors.

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So the first thing that we
need to do is build some

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vectors that describe
this triangle.

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And the vectors that we
need to build are

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P1, P2 and P1, P3.

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Since we're going to use these
in a minute, let's go ahead

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and compute them now.

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So P1, P2 is just the difference
of P2 minus P1.

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So we get a 0 minus
a negative 1.

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So we get 1, 2, and 1.

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Let me just check my notes to
make sure I did that right.

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Good.

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And P1, P3.

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We get again 0 minus
a negative 1.

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So 1, we get minus 1,
and then we get 1.

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Let me again check my notes.

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Very good.

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OK.

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So now that we have these
vectors, we need to remember

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that if we take the absolute
value of P1P2 cross product

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with P1P3, this will be equal
to the area of the

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parallelogram they enclose.

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So let's get started by
computing this cross product.

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So P1P2 cross P2P3.

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So remember to compute a cross
product, we take the

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determinant of a matrix where
we put in our unit normal

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vectors i, j, and k.

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And then we enter in, the
remaining entries of the

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matrix are just the entries
of our vectors.

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So we do 1, 2, 1.

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And 1, minus 1, 1.

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OK.

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And so we can compute this.

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And we get-- so the i component,
we get 2 minus a

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negative 1.

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So we get 3.

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Now the j component.

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If we look at the cofactor
matrix, it's just 1, 1, 1, 1,

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and that has determinant 0.

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So our middle component
is just 0.

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And finally the k component.

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We get minus 1, minus
another 2.

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So altogether, we get minus 3.

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So what that tells us now is
that this quantity here, the

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magnitude of the cross product,
is just 3 times the

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square root of 2, just
looking at the length

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of this vector here.

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So we're almost done, but let's
go back and look at what

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we had to start with.

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We were interested in the
triangle over here which was

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enclosed by the vectors P1, P2,
and the vectors P1, P3.

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And what we just computed is
actually the area of this

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parallelogram, which as you can
see is twice the area of

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the triangle that we're actually
interested in.

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So going back over here, we
see that the area of our

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triangle is equal to 3 root 2,
and we just need to divide by

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2 to get the triangle.

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OK?

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And I'll leave it at that.

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