Calculus III: The Dot Product (Level 8 of 12) | Scalar, Vector and Orthogonal Projections


The Dot Product level 8
In this video we will go over our first application
that makes use of the dot product. Recall
that we covered two different forms of the
dot product the component definition and geometric
definition. Having these two forms of the
dot product opens the possibility of using
basic algebraic operations like in the case
of the component definition to calculate geometric
quantities like lengths and angles via the
geometric definition.
In the previous video we used this to our
advantage to find the angle between two vectors
by rearranging the geometric definition. Another
useful application of the dot product is to
find the scalar and vector projections between
two vectors.
Many applications in physics and engineering
make use of vectors by adding two or more
vectors to produce a resultant vector. At
times the reverse problem, in this case decomposing
a given vector into the sum of two or more
vector components, is often encountered.
Given two vectors a and b with the same initial
point and theta representing the angle between
vector a and vector b, the scalar projection
of b onto a (also called the component of
vector b along vector a) is defined to be
the magnitude of the vector projection, you
can think of the scalar projection of vector
b onto vector a as the length of the shadow
that is cast when you shine a light in the
direction perpendicular or orthogonal to the
line that is parallel to vector a. Using right
triangle trigonometry we see that the scalar
projection is going to be equal to the magnitude
of vector b times cosine of theta. We mathematically
denote this length as follows, and is read
as the component of b along a. Similar to
the dot product this value will be positive
if theta is between 0 inclusive and pi over
2 exclusive, and it will be negative if theta
is between pi over 2 exclusive and pi inclusive.
Notice that this quantity appears in the geometric
definition of the dot product, using this
expression we see that the dot product between
vector a and vector b can be interpreted as
the length of vector a times the scalar projection
of vector b onto vector a.
The geometric definition of the dot product
can also be used to provide a simple way of
calculating the scalar projection of one vector
in the direction of another vector.
We know that the value of the scalar projection
is just equal to the magnitude of vector b
times cosine of theta, so we go ahead and
solve for this quantity, doing that we obtain
the following expression. Notice that the
component of vector b along vector a can be
computed by taking the dot product of vector
b with the unit vector in the direction of
vector a.
In other words, If you want to find out how
much a vector v “leans” in the direction of
a given vector you simply dot vector v with
a unit vector that points in the direction
of the given vector.
The scalar projection is nothing new, we have
already computed this quantity various times
when we broke apart a vector in R squared
into its x and y components and a vector in
R cubed when we broke it apart into its x,
y and z components. In these cases we found
the scalar projection of the vector in the
direction of the unit standard vectors i hat,
j hat in the case of R squared, and i hat,
j hat and k hat in the case of R cubed. What
makes the dot product so useful is the fact
that you can now easily calculate the component
of a vector in any direction and not just
the standard unit vectors i, j and k. The ability
to decompose a vector into its component parts
is a fundamental theme in physics, engineering
and linear algebra. We will see this in action
in a later video.
The vector projection of vector b onto vector
a (also known as the vector component of vector
b along vector a) can be computed by multiplying
the scalar projection of vector b onto vector
a with the unit vector in the direction of
vector a.
The vector projection will generate a vector
that points in the direction of the vector
that it was projected onto, with the length
of the shadow casted representing the scalar
projection. So keep this detail in mind the
scalar projection will generate a scalar that
represents the length of the shadow that is
created when one vector is projected onto
another vector, while the vector projection
will generate a vector whose magnitude is
equal to the length of the casted shadow.
In the special case where vector a is a unit
vector this expression simplifies to the following.
Personally I like to use the first form since
it makes more sense intuitively, than the
second expression, but you are free to use
either one they are essentially equivalent
expressions.
One last thing to note is that we need to
be very careful with our notation. The vector
projection of vector b onto vector a will
generate a totally different vector when compared
to the vector projection of vector a onto
vector b assuming that both vector a and vector
b are distinct vectors. In the first expression
the vector generated will be parallel to vector
a while in the second expression the vector
generated will be parallel to vector b, so
be careful with the notation and make sure
you double check that you are finding the
correct vector projection.
The final projection that you should be familiar
with is the vector component of vector b orthogonal
to vector a. This vector is denoted as follows,
this is referred to as the orthogonal projection
of vector b onto vector a, it is also known
as the vector rejection of vector b from vector
a, this vector is the orthogonal projection
of vector b onto a line that is orthogonal
to vector a. Both the vector projection and
vector rejection represent the components
of vector b. As a result the sum of the vector
rejection and vector projection is equal to
vector b. If we solve for the orthogonal projection
of vector b onto vector a it will be equal
to vector b minus the vector projection of
vector b onto vector a. Also notice that the
vector projection and vector rejection are
orthogonal to one another.
With these two projections you can essentially
break apart a vector into two components that
do not point in the direction of the standard
unit vectors i, j and k. So now you are free
to break a vector into components along any
vector. Alright in our next video we will
go over a couple of examples to illustrate
how to find these projections.

3 thoughts on “Calculus III: The Dot Product (Level 8 of 12) | Scalar, Vector and Orthogonal Projections

  1. The videos are great, but why there are very low views, may be you should consider changing the title, because so people are afraid of higher Mathematics courses like calculus 3,

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