Welcome to a video that will prove
the formula used to determine the angle
between two vectors.
Let's first review the formula.
If U and V are two vectors, then the angled
data between the two vectors can be defined by
cosine data is equal to U dotted with V divided
by the magnitude of U times the magnitude of V.
And before we start our proof, what we're going to do
is take a look at the sketch of vectors U and V.
So here's vector U in blue and vector V in red.
Now we want to determine V minus U or V plus the
opposite of U geometrically.
Remember to do that, we have to sketch the opposite
of U at the terminal point of vector V.
And we see that here as this thin blue vectors.
So vector V minus vector U is going to be
the vector with it's initial point
at the initial point of vector V and it's terminal point
at the terminal point of the opposite
of vector U as we see here.
What we're going to do now though is reorient these vectors
so that U and V share the same initial point and V
minus U would form the third side of a triangle.
So for our proof, we're going to be using this
triangle here.
And we're going to start by apply the law of cosines to
this triangle.
So again, here is the sketch of the triangle that
we're going to use, and here's the review of the
law of cosines.
And so we're going to let the magnitude of U minus V
be side C, so we'll have the magnitude of vector V
minus vector U squared must equal the magnitude of U
squared plus the magnitude of V squared, minus two
times the magnitude of U times the magnitude of V
times cosine C.
So cosine C represents the angle between the two vectors.
Let's go ahead and call this equation one.
And now we're going to introduce a dot product.
If we take the magnitude of V minus U squared
it's going to be the same as V minus U
dotted with V minus U.
Remember if we square this magnitude,
this squaring will undo the square root
and the result would be this dot product.
Now taking a look at the right side
again using the properties of dot products
this would be the same as V minus U
dotted with V, remember it's almost like distribution
and then we're going to have minus V minus U dotted with U.
Again that minus sign here
came from the subtraction sign here.
And now we'll go and expand this, this is going to be
V dotted with V minus U dotted with V,
then here we're going to have minus V dotted with U
then lastly we'll have plus U dotted with U.
Let's go ahead and recopy this last line on the
next slide and then continue.
Now we're going to go ahead and rewrite V dotted with V
as the magnitude of V squared
now remember the order of the dot plaque
doesn't matter so here we actually have minus two
times U dotted with V and then again we can
replace U dotted with U with the magnitude of U squared.
We'll go ahead and call this equation two.
Now what we're going to do is combine equation one and
equation two, let's go back and look at equation one.
Remember we started with the magnitude of U minus V
squared is equal to all of this, 'cause of the law
of cosines, so what we're going to do now is take the
right side of this equation, and set it equal to
the right side of equation two.
Square the magnitude of U squared plus the magnitude
of V squared minus two, a magnitude of U, magnitude of V
cosine C it's going to be equal to what we have here in blue.
So the magnitude of V squared minus two times U dotted
with V plus the magnitude of U squared.
This equation is going to simplify nicely,
notice we have magnitude of U squared on both sides
here and here those would simplify out
and the same thing with magnitude of V squared here
and the magnitude of V squared here.
So now if we can solve this equation for cosine C,
remember C is angled between the two vectors,
we should have our formula.
So last step is going to be to divide both sides by negative
two magnitude U, magnitude V.
Let's see what happens here.
So on the left side, notice it all,
this simplifies out we're left with cosine C.
And then on the right side the negative two's simplify out
and we have our formula, we have U dotted with V
all over the magnitude of U times the magnitude of V.
And there we have it and I've also provided some
examples of determining the angle C in another video.
I hope you found this proof helpful.