BAM!!! Mr. Tarrou. In this video we are going to take a look at the Dot Product and see
how it helps us find the angle between two vectors. Now the Dot Product is the process,
of well let's see, first you have to have two vectors in their horizontal and vertical
components. Here I have vector v which is a sub 1 times vector i plus b sub 1 times
vector j and vector w which is a sub 2 times vector i plus b sub 2 times vector j. I am
using these i and j notation, this horizontal component vector and the vertical component
vector which each has the length of one because my textbook stresses that a lot. You might
see this in a bit of a short hand version where it is just a,b in a set of pointed parenthesis
if you will. The Dot Product itself is found by taking the product, or multiplying the
a values a sub 1 times a sub 2 plus b sub 1 times b sub 2. You just this scaler multiple,
this numerical value. It alone does not necessarily mean a tremendous amount as it stands alone,
except for as you will see later and I will remind you now... That when the dot product
is equal to zero those two vectors are perpendicular. Now all this math operations, all of this
arithmetic is just dealing with addition and multiplication. Addition and multiplication
are commutative and associative so a lot of the properties that you learned way bay in
say algebra 1, a lot of those properties apply to Dot Products as well. The dot product is
associative...excuse me not associate but commutative. So it does not matter what order
you do the dot product in. If you are taking the dot product of a vector and the resultant
of two other vectors being added together, you can basically distribute that dot product
through the parenthesis and get vector u dot product of vector v plus vector u dot product
vector w. If you do the dot product between the zero vector and any other vector you will
get a value of zero. If you take the dot product between a vector and its self as you will
see in this example, that is the same as squaring the magnitude of that same vector and do actually
have to write that out. The square root that would see in the magnitude which is the distance
formula or the Pythagorean Theorem will cancel away with the power of 2. Why don't I just
show you that? So the magnitude of a vector if it is in component form is equal to the
square root of a squared plus b squared. Then it is the magnitude squared, so that is going
to cancel out that square root symbol giving you the a^2+b^2 I had a second ago. And we
have kind of like a combination of the associate and the commutative property a little bit
with the scalar multiple and the dot product. Let's go through an example. We have vector
v and vector w, and we want to find the dot product. We are going to take the two values
of 'a' and multiply them together. So 2 times this 'a' value of 1 and add that with the
b values of 4 times -1. 2 minus 4 is negative two. And that is it! You know. As it stands
alone, that just ends up being a scalar multiple of negative two and we are done. Now if I
take vector v and I do a dot product with itself and just say, ok let's take vector
v and find the dot product with vector v that is going to be the two 'a' values multiplied
together...so 2 times 2 plus the two 'b' values multiplied together and v is being dotted
with itself so I am using the same 'a' and the same 'b' values. So 4 times 4. Here I
have got 2 squared plus 4 squared. See what this is starting to look like? That is a^2+b^2
which is exactly what I had over here when I wrote out the magnitude formula. And that
ends up becoming 4 plus 16 which is equal to 20. So that is just a couple of quick examples
of dot products. It is just a little bit of multiplication and division. Let's see how
that helps us find the angle between two vectors. ALRIGHT! So here we have the formulas for
finding, there is two versions of it, we are going to have these formulas for helping us
find the angle between two vectors. These formulas, you probably have the derivation
in your textbook, they come from or are derived from the Law of Cosine. And it says that the
dot product between two vectors is equal to the magnitude of the first vector times the
magnitude of the second vector times the cosine of the angle that is in-between them. Now
if we solve that for theta, we are going to divide both sides by these magnitudes. Then
to get the cosine function away from theta we will have to do the inverse cosine. So
this is actually the format that you are going to be using most of the time which is just...
you are going to be given the vectors and then asked for the angle measure. Thus you
are probably going to be using the one that is set up to equal theta more often than the
one set up to equal the dot product. So here we have a couple of vectors and let's run
through this formula. Theta is equal to the inverse cosine of the dot product between
the two vectors...so we have 2(1)+4(-1)... that is going to be over the magnitude of
v which is going to be the square root of 'a' squared plus 'b' squared times the square
root of the magnitude of vector w... excuse me... times the square root of 1 squared plus
-1 squared or the magnitude of w. Now be aware that I wrapped my -1 with a set of parenthesis.
It might not be a bad habit, I usually do all of my substitutions wrapped in a set of
parenthesis just to make sure I don't make any sign errors. Because let's remember there
is a difference between -1^2 and (-1)^2. This comes out to be negative one, and this because
the negative is wrapped in the parenthesis is -1 times -1 which is positive one. This
just squares the 1. So let's not let our fancy calculators give a wrong answer because it
understands the difference in that notation. Now when you get all of this typed into your
calculator, there really is not much more to this depending on how many times or how
many steps you take getting this into your calculator, but it is going to come out to
be inverse cosine of -.316. This comes out to a theta value equal to 108.4 degrees. Now
let's take a look at this drawing really quickly and see why we can just that I am right from
the calculator right off the bat. Maybe we can have a little insight as to why this formula
is derived from the Law of Cosine. So let's get a different colored chalk here. Draw an
x y axis and put a few tick marks out here. Ok, so vector v says that it is 2,4... We
are going to the right two and up four and I am placing this on the origin only for convenience.
You do not need to have vectors always have their initial point on the origin. Vector
w is identifies i-j, so a is 1 and b is -1. That vector is right here. A rather short
one. We are finding the angle between those two vectors. This angle right here is what
we just figured out to be 108.4 degrees. Well, if we are looking for that angle between the
vectors we have enough information to find the magnitude of v. Indeed the magnitude of
v, that is 2^2 is 4, 4^2 is 16 so this length is the square root of 20. And this other vector
with just the a and b values of 1 and -1 has a length of square root two. So if you just
got done studying the Law of Cosine, you know that... now I don't have a full triangle set
up... but you knew that with the Law of Cosine that if you had side-angle-side... if you
had two sides of a triangle and the included angle, you would solve that triangle first,
or the first step you would be required to do was the Law of Cosine to figure it out.
There you can see that side-angle-side, and we are looking for the angle between those
two vectors. That set up is what we use or when we use the Law of Cosine when solving
oblique triangles. One more page. OH! One more comment. The 108.4. Remember that inverse
sine with your calculator will only give you answers between -90 and 90. It is the same
with tangent. But inverse cosine with your calculator will give you values between 0
and 180 degrees. And we are always going to be looking for the angle that the... the short
way around the circle. When vectors are opened up like this, we want the inside angle. We
do not necessarily need to wrap around the long way around those two vectors. If we need
to know that, we can just take this entire 360 degree rotation and subtract 108.4 from
it. So when you do this formula to find the angle between two vectors, you are going to
be finding an angle that is on the more closed side of those vectors... an angle that is
less than 180 degrees since that is what you get out of our calculator for the inverse
cosine since it has a restriction to act as an inverse cosine function. But, if I babble
on more I am going back to an old concept. Next screen! Well if we are finding the angle
between vectors, what does it look like when those vectors are perpendicular or parallel.
Well if they are perpendicular you can stop with the dot product. If your dot product
comes out to be zero, then those vectors are perpendicular or orthogonal as we like to
say in vector language. So what does it look like when they are parallel. Well when you
have vectors that are parallel, if the numbers are easy enough like these are you might notice
that they are scalar multiples of each other. 4 times 1.5 is 6 and 14 times 1.5 is 21. So
just the fact that I can see that these are scalar multiples of each other automatically
tells me that they are parallel, and all the signs are the same. So I also know that the
angle between those vectors is zero. Thus they are going to be parallel or going in
the same direction. If my signs had changed then I would have parallel vectors but going
in opposite directions. Well that is fine and dandy, but what if you can't just glance
at and see that they are scalar multiples of each other. You apply the formula. So we
have theta is equal to once again the inverse cosine of the dot product between the two
vectors. We are going to have 4 times 6 plus 14 times 21 over the magnitude of v...that
is going to be the square root of (4)^2+(14)^2... times the magnitude of vector v... the square
root of (6)^2+(21)^2. Now again you can work this out in tiny little pieces with you calculator
and show the many steps at a time. I am just going to give you the final answer. When you
get this typed into your calculator, you are going to get the value of.... oh! We don't
even need a calculator for this, right? I already noticed that these are scalar multiples
of each other and they are going the same direction. So I hope that when we get done
this answer comes out to be one. So theta equals the inverse cosine of 1, and won't
notice that to begin with maybe... but if you do notice this relationship you probably
are not trying to find the angle between them. And the inverse cosine of 1, well the inverse
cosine of 1. What do you put into an inverse trig function? You put into an inverse trig
function, the sides of a triangle and you get out an angle measurement. So around the
unit circle, where is your cosine value equal to one? The cosine value is equal to one,
your x/r value, is equal to one with a rotation of zero. So, you know, thus again as I noticed
ahead of time these vectors are parallel and they are going in the same direction. Whereas
if I made this negative four and negative fourteen, that would be -4 here and -14 there,
it would be -4 squared and I am running out of space here... plus a negative -14 squared.
My answer, these negatives are going to get squared away anyway and my top number instead
of being positive will become negative. Where are on the unit circle now is inverse cosine
of negative one? What rotation gives me cosine value of negative one? That is going to be
at -180 degrees. So that is just a last little step there for your special cases of when
your vectors are either perpendicular with a dot product of zero, or parallel. There
are two cases of being parallel. You might want to be able to remember or note which
it is. Are they going in the same direction or are they going in opposite directions.
I am Mr. Tarrou. BAM!!! Go do your homework:D