### Video Transcript

Suppose π΄π΅πΆπ· is a square of
side 47. Determine π΄π΅ dot π΄πΆ.

Letβs draw this square. Hereβs our square. And as this square is π΄π΅πΆπ·, the
vertices of the square must be π΄, π΅, πΆ, and π·. And more than that, the vertex π΄
must be adjacent to vertex π΅ which is adjacent to vertex πΆ which is adjacent to
vertex π·. So for example, we couldnβt swap
the labels of vertices π΅ and πΆ because then vertex π΅ would not be the next vertex
along from vertex π΄. Instead, weβd have the square
π΄πΆπ΅π·. And we should mark the length of
the sides on our diagram too. Our task is to determine the dot
product of the vectors π΄π΅ and π΄πΆ.

Hereβs the vector π΄π΅ which runs
along one of the sides of the square. And hereβs the vector π΄πΆ along
one of the diagonals. The question is, what is their dot
product? There are two ways we could solve
this problem. We could either use the geometric
definition of the dot product which involves the angle between the two vectors. Or, we could lay down a coordinate
system and use the formula in terms of the components of the two vectors. Letβs first use the geometric
definition of the dot product, that the dot product of two vectors is equal to the
product of their magnitudes times the cosine of the angle between them. We apply this definition to our
scenario. We know that the magnitude of the
vector π΄π΅ is just the length of side π΄π΅, which from the question we know is
47. The magnitude of π΄πΆ is less
obvious. But π΄π΅πΆπ· is a square. And so the triangle π΄π΅πΆ is a
right triangle of which the diagonal π΄πΆ is the hypotenuse. And so we can apply the Pythagorean
theorem to find that the length of the diagonal π΄πΆ, and hence the magnitude of the
vector π΄πΆ, is the square root of 47 squared plus 47 squared.

The only thing left to find is π,
the measure of the angle between the two vectors. We mark this in and itβs pretty
straightforward to see that the value of π is 45 degrees. If youβre not convinced, then you
can check that the diagonal π΄πΆ bisects the right angle at π΄, or you can consider
the angles in the triangle π΄π΅πΆ which is a right isosceles triangle. Anyway, we use this value of
π. And we now have an expression that
we could just put into our calculator if we wanted to. But itβs good practice to do this
by hand.

The factor of 47, we leave as it
is. Inside the square root, we have two
times 47 squared. And so we can take that factor of
47 out of the square root to give us 47 root two. As 45 degrees is a special angle,
we should remember the value of the cos 45 degrees. Itβs root two over two or
equivalently, one over square root two. The square root two and the one
over square root two cancel. And so weβre left with 47 times 47
which is 2209.

Now that we found π΄π΅ dot π΄πΆ
geometrically, letβs see another approach which involves setting down a coordinate
system. We can lay down a coordinate system
or more technically, specify a basis for our vectors by choosing directions for the
unit vectors π and π which weβre going to write our vectors in terms of. We can choose the unit vector π to
be parallel to the vector π΄π΅. And once weβve done that, we donβt
really have much choice about the direction of π, it has to be perpendicular to
π. You can either point up or down and
weβve chosen to make it point up. We can write our vectors π΄π΅ and
π΄πΆ in terms of these two vectors π and π. π΄π΅ has magnitude 47 and points in
the same direction as the unit vector π and so it must be 47π. π΄πΆ is slightly more difficult to
write in terms of π and π.

To help us, we find π΅πΆ in terms
of π and π. π΅πΆ has a magnitude of 47 and is
pointing in the opposite direction to the unit vector π. And so it is negative 47π. Back to π΄πΆ now, looking at the
diagram, we can see that π΄πΆ is π΄π΅ plus π΅πΆ. And because we have π΄π΅ and π΅πΆ
in terms of π and π, itβs now simple to write π΄πΆ in terms of π and π. Letβs clear away some of the
working before moving on. Remember that we wanted to find the
dot product of these two vectors. And so it might be helpful to first
write them in component form as the unit vector π has components one, zero. Our vector π΄π΅ which is 47π has
components 47, zero. And itβs a similar story for the
vector π΄πΆ. Here, weβve also used the fact that
the unit vector π has components zero, one.

Now weβre ready to find their dot
product. Itβs the product of the
π₯-components, so 47 times 47, plus the product of the π¦-components, thatβs zero
times negative 47. Here, weβre using the general rule
for the dot product of vectors in component form. Of course, zero times negative 47
is just zero. And so weβre left with 47 times 47,
as before. So the dot product of the vectors
π΄π΅ and π΄πΆ is 2209.

Here, weβve seen two methods of
solving this problem, using two ways of thinking about the dot product. The first where we used the
geometric definition of the dot product involving the magnitudes of the vectors and
the measure of the angle between them. And the second where we used the
formula for the dot product of vectors in component form. When tackling a problem about the
dot product, itβs a good idea to have both ways of thinking about the dot product in
mind. You may find that for a given
problem, one of the ways of thinking about the dot product gives an easier
solution.