DAVID JORDAN: Hello, and welcome back to recitation.
In this problem, I'd like you to compute
the area of a triangle.
This triangle sits in space and it has its three vertices,
labeled here as P1, P2, and P3.
So we're going to compute this area, and we're going to do it
using the cross product, which we learned about in lecture.
So why don't you take some time to work this out, pause
the video, and we'll check back in a minute and
see how I did it.
Hello and welcome back.
So the first thing that I like to do with a problem like this
is I like to draw a picture so I can kind of think about
what's going on.
So we have this triangle sitting out in space.
And we know that we want to take a cross product in order
to compute its area, but we need to be careful.
Cross product, it doesn't make sense to take
cross product of points.
What makes sense is to take cross product of vectors.
So the first thing that we need to do is build some
vectors that describe this triangle.
And the vectors that we need to build are
P1, P2 and P1, P3.
Since we're going to use these in a minute, let's go ahead
and compute them now.
So P1, P2 is just the difference of P2 minus P1.
So we get a 0 minus a negative 1.
So we get 1, 2, and 1.
Let me just check my notes to make sure I did that right.
Good.
And P1, P3.
We get again 0 minus a negative 1.
So 1, we get minus 1, and then we get 1.
Let me again check my notes.
Very good.
OK.
So now that we have these vectors, we need to remember
that if we take the absolute value of P1P2 cross product
with P1P3, this will be equal to the area of the
parallelogram they enclose.
So let's get started by computing this cross product.
So P1P2 cross P2P3.
So remember to compute a cross product, we take the
determinant of a matrix where we put in our unit normal
vectors i, j, and k.
And then we enter in, the remaining entries of the
matrix are just the entries of our vectors.
So we do 1, 2, 1.
And 1, minus 1, 1.
OK.
And so we can compute this.
And we get-- so the i component, we get 2 minus a
negative 1.
So we get 3.
Now the j component.
If we look at the cofactor matrix, it's just 1, 1, 1, 1,
and that has determinant 0.
So our middle component is just 0.
And finally the k component.
We get minus 1, minus another 2.
So altogether, we get minus 3.
So what that tells us now is that this quantity here, the
magnitude of the cross product, is just 3 times the
square root of 2, just looking at the length
of this vector here.
So we're almost done, but let's go back and look at what
we had to start with.
We were interested in the triangle over here which was
enclosed by the vectors P1, P2, and the vectors P1, P3.
And what we just computed is actually the area of this
parallelogram, which as you can see is twice the area of
the triangle that we're actually interested in.
So going back over here, we see that the area of our
triangle is equal to 3 root 2, and we just need to divide by
2 to get the triangle.
OK?
And I'll leave it at that.