The next topic for discussion is that of the dot
product. Let’s jump right into the
definition of the dot product. Given the
two vectors 
and the dot product is,
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(1)
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Sometimes the dot product is called the scalar product. The dot
product is also an example of an inner product and so
on occasion you may hear it called an inner product.
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Example 1 Compute
the dot product for each of the following.
(a)
(b)
Solution
Not much to do with these other than use the formula.
(a)
(b)
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Here are some properties of the dot product.
Properties
The proofs of these properties are mostly “computational”
proofs and so we’re only going to do a couple of them and leave the rest to you
to prove.
Proof of
Proof of : If
then
There is also a nice geometric interpretation to the dot
product. First suppose that θ is the angle between and such that as shown in the image below.
We can then have the following theorem.
Theorem
Proof
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Let’s give a modified version of the sketch above.
The three vectors above form the triangle AOB and note that the length of each
side is nothing more than the magnitude of the vector forming that side.
The Law of Cosines tells us that,
Also using the properties of dot products we can write the
left side as,
Our original
equation is then,
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The formula from this theorem is often used not to
compute a dot product but instead to find the angle between two vectors. Note as well that while the sketch of the two
vectors in the proof is for two dimensional vectors the theorem is valid for
vectors of any dimension (as long as they have the same dimension of course).
Let’s see an example of this.
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Example 2 Determine
the angle between and .
Solution
We will need the dot product as well as the magnitudes of
each vector.
The angle is then,
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The dot product gives us a very nice method for determining
if two vectors are perpendicular and it will give another method for
determining when two vectors are parallel.
Note as well that often we will use the term orthogonal in place of perpendicular.
Now, if two vectors are orthogonal then we know that the
angle between them is 90 degrees. From (2)
this tells us that if two vectors are orthogonal then,
Likewise, if two vectors are parallel then the angle between
them is either 0 degrees (pointing in the same direction) or 180 degrees
(pointing in the opposite direction).
Once again using (2) this would mean that one
of the following would have to be true.
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Example 3 Determine
if the following vectors are parallel, orthogonal, or neither.
(a)
(b)
Solution
(a) First get
the dot product to see if they are orthogonal.
The two vectors are orthogonal.
(b) Again,
let’s get the dot product first.
So, they aren’t orthogonal. Let’s get the magnitudes and see if they
are parallel.
Now, notice that,
So, the two vectors are parallel.
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There are several nice applications of the dot product as
well that we should look at.
Projections
The best way to understand projections is to see a couple of
sketches. So, given two vectors and we want to determine the projection of onto . The projection is denoted by . Here are a couple of sketches illustrating
the projection.
So, to get the projection of onto we drop straight down from the end of until we hit (and form a right angle)
with the line that is parallel to . The projection is then the vector that is
parallel to ,
starts at the same point both of the original vectors started at and ends where
the dashed line hits the line parallel to .
There is an nice formula for finding the projection of onto . Here it is,
Note that we also need to be very careful with notation
here. The projection of onto is given by
We can see that this will be a totally different
vector. This vector is parallel to ,
while is parallel to . So, be careful with notation and make sure
you are finding the correct projection.
Here’s an example.
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Example 4 Determine
the projection of onto .
Solution
We need the dot product and the magnitude of .
The projection is then,
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For comparison purposes let’s do it the other way around as
well.
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Example 5 Determine
the projection of onto .
Solution
We need the dot product and the magnitude of .
The projection is then,
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As we can see from the previous two examples the two
projections are different so be careful.
Direction Cosines
This application of the dot product requires that we be in
three dimensional space unlike all the other applications we’ve looked at to
this point.
Let’s start with a vector, ,
in three dimensional space. This vector
will form angles with the x-axis (α ), the y-axis
(β ), and the z-axis (γ ). These angles are called direction angles and the cosines of these angles are called direction cosines.
Here is a sketch of a vector and the direction angles.
The formulas for the direction cosines are,
Let’s verify the first dot product above. We’ll leave the rest to you to verify.
Here are a couple of nice facts about the direction cosines.
- The
vector is a unit vector.
-
-
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Let’s do a quick example involving direction cosines.
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Example 6 Determine
the direction cosines and direction angles for .
Solution
We will need the magnitude of the vector.
The direction cosines and angles are then,
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