# 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

start with two planar vectors A and B with

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.

Excellent videos! Thanks alot!

This makes so much sense geometrically now that I've seen this video.

How much of A contributes to B. And how much of B contributes to A.

So when they are perpendicular…none of A is in B = 0

Makes so much sense geometrically.

How incredibly clear this is!!!! You are an amazing teacher. Methodical, step-by-step and taking nothing for granted concerning prior knowledge. EXCELLENT!!!!

It has also demonstrated to me…after watching a number of videos by others on the dot product, just how mediocre and confusing most teachers are. My opinion.

Your video is very helpful. Thanks for help. God bless you.

awesome, thank you so much!

1:47 Scalar Multiplication Review

2:16 Dot Product (Scalar Product)

Terrific lecture! Thank you for sharing additional information on the dot product of two vectors particularly the 5 dot product cases.

Superb explanation….. Nice…. Keep it up sir…. Thanks for the video….