Welcome to the first of several videos
on vector cross products.
The goals of this video are to:
determine the cross product of two vectors in R3,
verify the cross product of two vectors
is orthogonal to the given vectors
and then also to determine the angle between two vectors.
To determine the cross product of two vectors we can
evaluate a three by three determinant where the first row
would be the i, j, k vectors, the second row would consist
of the components of vector u
and the third row would be equal
to the components of vector v.
And there are several ways to evaluate a three by three
determinant but we'll use the co factor expansion method,
meaning we'll first eliminate row one and column one.
The remaining elements here will form the two by two
determinant that we see here and we'll multiply that
by vector i.
And then we'll subtract the determinant form by eliminating
row one and column two.
Eliminate row one and column two, notice we'll have u one,
u three, v one, v three which we see here.
We'll multiply that by the j vector
and then we'll add the value of the determinant form
by eliminating row one and column three.
So we see here, u one, u two, v one, v two
and then multiplying that by vector k.
Now what's special about the cross product of two vectors
is the result is another vector that is orthogonal
or perpendicular to both vector u and vector v.
Let's take a look at this.
Here's an illustration where we have the cross product
of the red and blue vector and what I want you to notice is
that the black vector or the cross product is perpendicular
to the red vector as we see here as well as,
being perpendicular to the blue vector as we see here.
So the black vector is the cross product of u and v.
Now you might be wondering "why is it that the cross product
of the black vector points in this direction other than
the opposite direction?".
And that's an important thing to consider
and before we take a look at this in more detail
I want you to be careful not to confuse the cross product
of two vectors with the dot product of two vectors.
Remember the dot product results in a scalar or numerical
value however, when we cross two vectors,
we get another vector that is perpendicular to both.
Okay so let's talk about how we determine what direction
the cross product vector will go.
This is called the right hand rule, what we're going to do
is take our right hand and point our fingers
in the direction of vector u
so that when we close our hand or bend our fingers,
they bend towards vector v.
So for example, in this diagram here if we point our fingers
in the direction of the u vector with our thumb pointing up
if we bend our fingers they will bend toward vector v.
Our thumb points up so the orthogonal vector
will also point up.
So the order in which we cross these two vectors
is important, we point our fingers first in the direction
of u so that they fold or bend toward v.
If we switch the orientation of the two vectors,
if we want our fingers to close toward vector v
we'd have to have our hand almost upside down
so that our thumb points down so that if we bend our fingers
they would bend towards vector v, thumb would point down
therefore, the orthogonal vector would point down.
So this is the right hand rule and this is how we determine
what direction the orthogonal vector will go.
Let's take a look at our example now.
We want to determine the vector that is orthogonal
to both u and v, so we'll have to find the cross product.
So we'll have u crossed with v is going to be equal
to a three by three determinant where the first row consists
of the i, j and k vectors.
The second row consists of the components of vector u,
so we have three, one, negative five.
Third row consists of the components of vector v.
Now we'll go ahead and evaluate this, the first two by two
determinant will consist of the elements remaining after
eliminating row one and column one.
So we'll have one, negative five, negative two,
negative three.
Going to multiply this by i vector
and then we're going to subtract the two by two determinant
form by eliminating row one and column two.
So we'll have three, negative five, four, negative three
times vector j, plus,
now we're going to eliminate row one and column three,
so we'll have three, one, four, negative two
times vector k.
Let's go ahead evaluate this.
We're going to have negative three minus positive ten,
that's going to mean negative 13 times vector i, minus,
here we'll have negative nine minus negative 20
which will become negative nine plus 20 with positive 11
but we do have this minus sign here so I'll minus 11j
plus, here we'll have negative six minus four,
that's negative 10 times k.
So instead of writing plus negative 10 I'll go ahead
and just right minus 10k.
So this vector should be orthogonal to both vector u
and vector v.
Let's go and take a look at the graph of this.
It's a little hard to interpret this
because we're looking at a three dimensional graph
on a two dimensional surface but you can see that
this blue vector here, the cross of u and v is perpendicular
to both the red and brown vectors as we see here.
Now that we have found u cross of v,
let's verify that this vector is orthogonal to both u and v.
Remember the way we can verify this is if we dot u
and the cross product and v and the cross product,
if that dot product is equal to zero,
those vectors are perpendicular.
So we have to calculate two dot product,
first we have to dot u with our cross product
to make sure this is equal to zero.
Then we have to dot v,
with the cross product.
So let's go ahead and verify this.
Here we'd have three times negative 13, that's negative 39
plus one times negative 11, that's negative 11
plus negative five times negative 10, that's 50.
Well this is negative 50 plus 50 so that does equal zero.
So these two vectors are perpendicular.
Let's go ahead and check to make sure the cross product
is perpendicular with vector v.
Here we'd have four times negative 13, that's negative 52
plus negative two times negative 11, that's 22
plus negative three times negative 10 is 30
and again this is equal to zero.
So we have verified that the cross product is orthogonal
to both u and v.
Before we move on let's go ahead
and talk about some additional properties of cross products.
The magnitude of u cross with v is equal to the magnitude
of u times the magnitude of v times sine theta,
where sine theta is the angle between the two vectors.
Number two, u cross v equals zero if and only if
the two vectors are scalar multiples of each other.
And then lastly, the magnitude of the cross product
equals the area of the parallelogram having u and v
as adjacent sides.
Now we'll talk more about these in future videos
but let's go and take a look at one example
of this first property.
Going back to the same problem we can use this property
to determine the angle between our given vectors u and v.
Let's go ahead and try to do that.
If we solve this for sine theta we'd have sine theta
must equal the magnitude of u cross v
divided by the magnitude of u times the magnitude of v.
Once we find this we can determine the angle
between the two vectors, let's give it a try.
So here's u cross v we need to find the magnitude of this.
Going to have the square root of negative 13 squared
plus negative 11 squared, plus negative 10 squared.
Our denominator's going to consist of two different
square roots.
First, a magnitude of u square root of three squared
plus one squared, plus negative five squared,
multiply this by the magnitude of v.
So we'll have four squared plus negative two squared
plus negative three squared.
Now to save a little bit of time
I've already simplified this.
This comes out to approximately
19 point seven four eight four
divided by approximately 31 point eight five nine one.
Now using some trig, if sine theta is equal to this ratio
that means arc sine of this ratio
would give us an angle theta.
So let's go ahead and get our calculators out
and give that a try.
Let's first put our calculator in degree mode,
notice I'm already in degree mode here.
So now we're going to press second sine
for arc sine or inverse sine
then we'll just type in our ratio.
So the angle between those two vectors is approximately
38 point three degrees.
Okay, that's going to do it for this first video.
Thank you for watching.