Hi everyone.
Welcome back to integralcalc.com.
Today we're going to talk about how to find the cross product of 2 vectors.
The vectors that we're given in this problem are a equals 5, negative 1 negative 2 and
b equals negative 3, 2 and 4.
So these are not in component form.
The numbers that were given here are not coefficients on i, j and k.
We can just take these 2 vectors and go straight into finding the cross product.
The formula that we’re going to need is just this strange formula.
It's a grid, a third order determinate grid of positives and negatives that tells us what
signs we're going to put on our second order determinates.
You’ll see how it works in just a second.
This problem isn't too hard especially if you're given two vectors like this with the
numbers already broken out for you here.
But the first thing we're going to do when we have a problem like this and we're given
these two vectors is we build a third order determinate matrix here.
We're always going to put i, j, and k in the first row in this order, so go ahead and write
i, j and k out in the first row.
And then we're going to take our direction numbers here in a, a1, a2, and a3 and we're
going to put them here on our second row underneath i, j, and k so 5, negative 1 and negative
2.
And then we take b1, b2 and b3 and put them in our third row as negative 3, 2, and 4 here.
And that's how we do it.
That kind of sets our problem for us.
So it's always i, j, and k then a then b in the first, second and third row.
And now that we've filled this third order determinate, we can go ahead and break that
out into second order determinates.
So here’s what that's going to look like.
And this is the way that we break it out.
First of all, you'll notice that we have i, j and k here multiplied out with our second
order determinates.
That's always going to be the case.
You can go ahead and put i, j and k out in the front, then we need to remember and this
is the reason that we have this formula up here; we're always going to work with this
formula.
We start with positive on the upper left-hand corner and no matter how big your matrix gets,
you follow this checker board pattern of positive negative positive and then you start the second
row with a negative.
Just make a checkerboard of positive signs and then fill the rest with negatives.
But what this tells you is that we're going to assign a positive sign to i, which is the
variable we have here on the upper left-hand corner of this matrix.
And then a negative value to, in our case j here which is why this j here is negative.
And then k of course is positive because k is in this position here and when we look
at our formula and we follow this checkerboard pattern, k is a positive based on this right
here.
So that's why we get positive i, negative j and positive k.
And no matter how big your matrix is, you can follow that checkerboard pattern.
Now as for the second order determinates here, the way that we're going to do it is the following.
For i, what we have multiplied by i, we're going to take our third order determinate,
look at it and we're going to eliminate everything that's in the same row and the same column
as i.
So in our case, when we're looking at our third order determinate, we're going to eliminate
5 and negative 3 because they are in the same column as i.
So we’re going to ignore them.
And then we're also going to ignore j and k because they are in the same row as i.
So we block out everything in this row and everything in this column because that is
where i is and we’re only going to look at what remains which in our case is negative
1, negative 2, 2 and 4.
And so you see, that's why those four numbers come over here to this second order determinate
which gets multiplied by i.
That negative 1, negative 2, 2 and 4.
And then we're going to do the same thing for j and k.
So let's just look at j and practice one more time.
For j we're going to eliminate everything in the row with j, so i, j and k here are
ignored completely.
And then we're going to eliminate everything in the column with j.
So in our case here, we're going to ignore negative 1 and 2.
Everything here get's ignored.
And everything else gets translated in the same orientation on the second order determinate.
You don't mix around the numbers at all.
You just move them in their position right over.
So we have left 5, negative 3, negative 2 and 4.
And so they move over in that order.
5 being in the upper left-hand corner, negative 2 being in the upper left hand corner negative
3 on the lower left and 4 in the lower right.
And then we do the same thing with k, ignoring the row i and j and ignoring the column negative
2 and 4 so we end up with everything left.
5, negative 1 negative 3 and 2.
So that's how we build out our second order determinates.
And the reason that we do that is because once we have, we're going to do some multiplication
that allow us to simplify and then we're going to get our final answer for the cross product.
So the way that we do that is let's look at i here.
We're going to have i times whatever we've got in the second order determinate and what
we do here is we multiply what's in the upper left-hand corner by what's in the lower right-hand
corner.
So we do negative 1 times 4, which is going to give us a negative 4 and that goes in the
first spot here, negative 4.
Then we're always going to subtract.
We’re always going to subtract and then we multiply the lower left and the upper right.
So, 2 times negative 2 is a negative 4 so we have negative 4 here in the second spot.
So upper left times lower right minus lower left times upper right and we follow that
same pattern.
So you’ll see with j here, we'll bring this negative j down from here and we'll get upper
left times lower right, it's a diagonal pattern here, five times 4 is 20 so we get 20 on the
first spot, always subtract and then negative 3 times 2 on the diagonal again is a positive
6.
So we're going to get 20 minus a positive 6.
And we're going to do the same thing with k.
So we build that out and once we've done it, all we have to do is simplify.
So we'll get negative 4 minus a negative 4 is going to be a negative 4 plus 4, which
will give us 0, 20 minus 6 gives us 14 and 10 minus 3 gives us 7 here.
So we simplify and then we'll do one more simplification step where we just make these
coefficients.
So we've got 0i minus 14 j plus 7k is as simple as we can get.
And the reason that I left out this 0i is because these coefficients now, 0, negative
14 and 7 are going to be with 3 direction numbers that are going to compose our cross
product factor.
So our final answer is going to be the vector 0, negative 14 and 7.
And that ends up being our final answer for our cross product, which we just pulled directly
from here, these coefficients on i, j and k, 0 negative 14 and 7.
So I hope that helped you guys.
Not too bad for cross product.
And I will see you in the next video.
Bye.