Cross Product Example 2


Uploaded by TheIntegralCALC on 16.01.2011

Transcript:
Hi everyone. Welcome back to integralcalc.com. Today we're going to be doing another example
of how to take the cross product of two component vectors. Our vectors in this case are in component
form and we've been given a equals i minus j plus 3k and b equals negative 2 i plus 3j
plus k. So we're given vectors in component form.
The only formula we only need to remember is the one here. Which tells us that when
we break this component vectors when we take the third order determinate matrix and then
break them up into second order determinates, we're going to be using this positive-negative
pattern here to decide the sign, positive or negative of our second order determinates.
So you'll see how that works when we get to the next step.
So the first thing that we're going to do when we've been given these component form
vectors is to create a third order determinate matrix here which just mean that we've got
a 3x3 matrix. And we're always going to put i, j and k in that order in the first row
of the matrix. And then for the second and third rows, we're
going to take the coefficients from our original vectors a and b. Sometimes, you're given like
a point with direction numbers in your vector and you're just given 1, negative 1 and 3.
In this case, all we have to do is look at the coefficients here on a. So you see that
the coefficient on i is 1. It’s implied which is why it's invisible. We have a negative
1 on the j because it’s a negative here and then a 3 as a coefficient on our variable
k. So that means that our direction numbers for a are 1 negative 1 and 3.
So a always goes in the first row underneath i, j and k and we just take those coefficients
and put them in order right underneath there. And then we find the coefficients in b and
make that our third row. So in this case we have -2, 3 and 1. So we put down here -2,
3, and 1. And that's how you build our third order determinate
matrix. And then from there, we're going to take it and break it apart into second order
determinate pieces so that we can simplify and solve for the cross product. So here's
what that's going to look like. The first thing that we're going to do, you'll
notice that we have i, j, and k multiplied by this second order determinate matrices
so we're always going to have that. Notice that i is positive, j is negative and k is
positive. Well, the reason that we have that is because i here , in our third order determinate
matrix is in the upper left-hand corner, right? Well, if we're following our formula, we have
a positive here in the upper left-hand corner which is why i is positive. j is negative
because it's in the second spot right here and our formula tells us that that spot is
supposed to be negative, so we have a negative j. And then a positive k because k here corresponds
to this positive sign here. And it doesn't matter what your original matrix
is. So if you're dealing with cross products and you build a 2x2 matrix, a 3x3 matrix,
a 7x7 matrix it doesn't matter. You're always going to start with a positive sign in the
uppermost left-hand corner and then build like a checkerboard pattern alternating positive,
negative, positive, negative and then starting the second row with negative positive, negative,
positive. So you build that checkerboard pattern and then that's how you assign to i j and
k here. So just keep that in mind. And then as far as these second order determinate
matrices go, the way that we build them is the following. The one here that's multiplied
here by i, what we do is we look at our third order determinate matrix, the original one
that we built here and we're going to ice i in the first position here. We're going
to ignore everything in the same column and the same row as i. So we're going to ignore
j and k because they're in the same row with i and we're also going to ignore 1 and  -2
here because they're in the same column with i. So we're ignoring this column and we're
ignoring this row which means that we take everything that's left and make that our new
matrix that we multiply by i and everything that's left in that case is -1, 3, 3, and
1 which is why we put the -1, 3, 3, and 1 her and they're always going to translate
in the exact same position. So you're just going to take that whole block and move it
straight over. You’re not mixing up the numbers in any way.
So we'll just look at what we multiplied here by j. Remember that we're ignoring everything
in the same row as j so we ignore i, j and k here, ignoring the row. And then we're also
going to ignore the column so we ignore negative 1 and 3 because they're in the same column
as j, which mean that we're left with 1, 3, -2 and 1 and we're going to take that 1, 3,
-2, and 1 and move it straight over and we get that 1, 3, -2 and 1 multiplied by j, again
not mixing up the numbers. We do the same thing with k.
So once we've done that, we go ahead and start simplifying to solve the cross product and
the way we do that is we keep i, j and k, exactly as they are. We’ve got positive
i, negative j and positive k here. And then we get these number by looking at this matrix
that's multiplied here by i. What we're going to do is multiply the upper left number by
the bottom right number so -1 times 1 is -1. So we get -1 here on this first spot. That’s
how we get that number. We’re always going to subtract in the middle. You’re always
going to subtract. And we get the second number, in this case 9 by multiplying the bottom left
number with the upper right number. So 3 times 3 is 9, which is why we get the 9 here.
So again, it's a diagonal pattern. -1 times 1 here minus 3 times 3 this way and we're
going to do the same thing with j. so 1 times 1 here gives us 1 in this spot, we always
subtract and then we do -2 times 3 here which give us a negative 6 so we subtract that negative
6 there. Don’t forget to include both negatives because it's always going to be subtraction
and you might get a negative sign and get that double negative.
So that's how we build that step here. And then our final step is just simplification.
We’re going to obviously simplify -1 minus 9 is -10, 1 minus a -6 which is going to be
1 plus 6 which gives us 7. And then 3 minus 2 on the k is 1.
And then I'll go ahead and simplify one more step and make these three coefficients, our
i, j and k for simplicity so we're going to end up with -10i minus 7j plus k.
And once we've done that, to find our cross product vector, all we do is take the coefficients
on i, j and k in this new vector and put them into these point from here so we're going
to have -10, -7 and 1 point from the -10, -7, and 1 coefficients here on this vector.
This is the component form of the vector just like we had here on our original problem and
then this is just the point itself. So -10, -7, 1 is our cross product for these two vectors
in component form. So I hope that helped you guys and I’ll
see you in the next video. Bye.