# Calculus III: The Dot Product (Level 1 of 12) | Geometric Definition

The Dot Product Level 1
In this video we will define a new vector
operation called the dot product also known
as the scalar product and sometimes referred
to as the inner product. Defining vector multiplication
is pretty tricky because, vectors are not
ordinary numbers, so ordinary multiplication
is not directly applicable to vectors. Vectors
have direction as well as magnitude, they
cannot be multiplied in the same way that
scalars are. Instead we must define what the
operation of vector multiplication means.
Among the many possible ways to define how
to multiply vectors, there are three ways
that is most commonly used in math and physics.
These include multiplication of a vector by
a scalar, which was already discussed in the
previous videos. Multiplication of one vector
by a second vector so as to produce a scalar,
and multiplication of one vector by a second
vector so as to produce another vector. In
the following series of videos we will focus
on the second type, known as the
dot product (because a dot is used to indicate
multiplication). In these videos I will be
using the dot product and scalar product interchangeably.
The most common use of the dot product in
applications in physics and engineering is
to decompose vectors into their components
parallel and perpendicular to a given vector.
Let’s first go over a visual demonstration
of how the dot product is defined geometrically.
In the previous videos we described scalar
multiplication as an operation that takes
a real number called a scalar and changes
the length or magnitude of a vector, if this
scalar was negative then it would also change
the direction of the vector 180 degrees and
would point in the opposite direction. This
operation can be denoted as c times vector
v .
Now the scalar product is different because
it involves multiplication of 2 vectors as
oppose to multiplication of a scalar and a
vector. So keep this little detail in mind.
The next difference between scalar multiplication
and the scalar product is that scalar multiplication
produces a vector and the scalar product produces
a scalar not a vector.
To define the scalar product we will
known magnitudes and place them tail to tail
as follows. Where the angle theta between
the vectors can range anywhere from 0 degrees
to 180 degrees.
Let’s take a look at the first case when
the angle between vector A and vector B is
equal to 0 degrees in other words they are
parallel. Under this condition the product
of two vectors behaves similar to the way
we multiply real numbers for example if vector
A has a length or magnitude of 10 and vector
B has a magnitude of 4 then the scalar product
pronounced as A dot B would be equal to 10
times 4 which simplifies to 40. Now the only
reason why we were able to multiply the magnitudes
was because both vectors were pointing in
the same direction so their magnitudes multiply
like real numbers. Unfortunately vectors are
rarely lined up and point in the same direction.
So know we need the scalar product to account
for potential differences in direction.
Let’s take a look at a different case.
What if the angle between these same vectors
was an acute angle? Now how do we define the
scalar product? Well, when the vectors lined
up in the case when the angle was 0 degrees
all we had to do was to multiply the magnitudes,
so we essentially need to do the same for
this case. The only problem is that vector
B is not pointing in the same direction as
vector A, but relative to an xy-coordinate
plane the x-component of vector B does just
that and points in the same direction or along
the path of vector A, this is nothing more
than the projection of vector B in the direction
of A. In other words some of vector B is projected
onto vector A. We now have a way to take the
direction of the vectors into account, in
this sense the dot product represents the
product of the magnitude of one vector and
the magnitude of the other vector along the
direction of the first vector.
Using right triangle trigonometry we can easily
find the projection of vector B onto vector
A by computing the x-component of vector B
in this case it will be equal to the magnitude
of vector B times cosine of theta, this expression
represents the component of vector B along
the direction of vector A we now have a vector
that points along the path of vector A so
we take this value and multiply it by the
magnitude of vector A to find value of the
scalar product, once again notice that we
end up with a numerical value, a scalar and
not a vector this is really important, the
scalar product produces a scalar not a vector.
Now what if we wanted to do the opposite and
find the projection of vector A in the direction
of vector B? Well it turns out that we can
take components along any direction that’s
convenient, not just the x and y-axes. So
we go ahead and draw a line that points in
the direction of vector B and draw a line
perpendicular to this line that stops at the
tip of vector A. Once again we can use right
triangle trigonometry to find the projection
of vector A along the direction of vector
B in this case it will be equal to the magnitude
of vector A times cosine of theta. Now that
we have a vector that points in the direction
of vector B it is just a matter of multiplying
this expression by the magnitude of vector
B. Notice that in both cases the final expression
is the same. As an example let’s find the
dot product between vector A and vector B
when theta equals 60 degrees. Using the geometric
interpretation of the dot product we go ahead
and multiply the magnitudes together obtaining
40 and then multiply by cosine of 60 degrees
which is equal to one half, simplifying the
expression we obtain 20 as our final answer.
Notice that the dot product was reduced in
half when compared to the first case when
both vectors pointed in the same direction
where theta equaled 0 degrees. This occurred
because the projection of one vector in the
direction of the other vector was reduced.
Even though both vectors had the same magnitude
the direction changed the value of the final
answer. This tells us that the dot product
is essentially multiplication taking direction
into account.
Now let’s take a look at the case when theta
equals 90 degrees or pi over 2. Under this
condition neither of the vectors has a projection
in the direction of the other vector. So we
are essentially multiplying the magnitude
of the one vector times zero which simplifies
to 0. This result tells us that the scalar
product of two perpendicular vectors or orthogonal
vectors is always equal to zero. In some textbooks
the words perpendicular, orthogonal, and normal
all mean essentially the same thing in this
case meeting at right angles. However, it
is common to say that two vectors are orthogonal,
two lines or planes are perpendicular, and
a vector is said to be normal to a given line
or plane.
So 2 vectors A and B are orthogonal if the
dot product is equal to 0. From this definition,
it follows that the zero vector is orthogonal
to every vector v, because zero dot v equals
0, moreover, for any angle theta between 0
and pi , cosine of theta equals 0 if and only
if theta equals pi over 2. So two nonzero
vectors are orthogonal if and only if the
angle between them is pi over 2. This fact
will play a crucial role in later topics in
this course.
Now, what happens when the angle between the
vectors is obtuse? In this case the geometric
interpretation is the same we still want to
find the projection of one vector in the direction
of the other vector and multiply it by the
magnitude of the second vector. In this case
some of vector B is projected onto the negative
of vector A, also the product between the
magnitudes will be the same since magnitudes
are always positive, but the angle is obtuse
meaning that cosine of theta will be negative.
For example if theta equals 120 degrees A
dot B will be equal to negative 20.
Finally, if theta equals 180 degrees then
all of vector B is projected onto the negative
of vector A, in essence both vectors are parallel
in opposite directions or in this case anti-parallel.
This is very similar to the first case where
both vectors were parallel and pointing in
the same direction, in this case the dot product
will have the same value as case 1 but it
will be negative.
Alright notice that the dot product is extremely
useful if you are interested in finding out
how much of one vector is projected onto the
second vector or how similar 2 vectors are
in direction. Taking a look at all 5 cases
we see that if the magnitudes of 2 vectors
are constant and do not change the dot product
will have a maximum value when theta equals
0 degrees and will have a minimum value when
theta equals pi. If theta is an angle between
0 and pi then the dot product will have some
fractional portion of the maximum value if
theta is acute and some fractional portion
of the minimum value if theta is obtuse. In
the special case when theta equals pi over
2 the dot product will be equal to 0 and the
vectors will be orthogonal.
Aright in our next video we will calculate
the dot product by using the component definition.

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