Video Transcript
Suppose 𝐴𝐵𝐶𝐷 is a square of
side 47. Determine 𝐴𝐵 dot 𝐴𝐶.
Let’s draw this square. Here’s our square. And as this square is 𝐴𝐵𝐶𝐷, the
vertices of the square must be 𝐴, 𝐵, 𝐶, and 𝐷. And more than that, the vertex 𝐴
must be adjacent to vertex 𝐵 which is adjacent to vertex 𝐶 which is adjacent to
vertex 𝐷. So for example, we couldn’t swap
the labels of vertices 𝐵 and 𝐶 because then vertex 𝐵 would not be the next vertex
along from vertex 𝐴. Instead, we’d have the square
𝐴𝐶𝐵𝐷. And we should mark the length of
the sides on our diagram too. Our task is to determine the dot
product of the vectors 𝐴𝐵 and 𝐴𝐶.
Here’s the vector 𝐴𝐵 which runs
along one of the sides of the square. And here’s the vector 𝐴𝐶 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. Let’s 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 𝐴𝐵 is just the length of side 𝐴𝐵, which from the question we know is
47. The magnitude of 𝐴𝐶 is less
obvious. But 𝐴𝐵𝐶𝐷 is a square. And so the triangle 𝐴𝐵𝐶 is a
right triangle of which the diagonal 𝐴𝐶 is the hypotenuse. And so we can apply the Pythagorean
theorem to find that the length of the diagonal 𝐴𝐶, and hence the magnitude of the
vector 𝐴𝐶, is the square root of 47 squared plus 47 squared.
The only thing left to find is 𝜃,
the measure of the angle between the two vectors. We mark this in and it’s pretty
straightforward to see that the value of 𝜃 is 45 degrees. If you’re not convinced, then you
can check that the diagonal 𝐴𝐶 bisects the right angle at 𝐴, or you can consider
the angles in the triangle 𝐴𝐵𝐶 which is a right isosceles triangle. Anyway, we use this value of
𝜃. And we now have an expression that
we could just put into our calculator if we wanted to. But it’s 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. It’s 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 we’re left with 47 times 47
which is 2209.
Now that we found 𝐴𝐵 dot 𝐴𝐶
geometrically, let’s 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 𝑖 and 𝑗 which we’re going to write our vectors in terms of. We can choose the unit vector 𝑖 to
be parallel to the vector 𝐴𝐵. And once we’ve done that, we don’t
really have much choice about the direction of 𝑗, it has to be perpendicular to
𝑖. You can either point up or down and
we’ve chosen to make it point up. We can write our vectors 𝐴𝐵 and
𝐴𝐶 in terms of these two vectors 𝑖 and 𝑗. 𝐴𝐵 has magnitude 47 and points in
the same direction as the unit vector 𝑖 and so it must be 47𝑖. 𝐴𝐶 is slightly more difficult to
write in terms of 𝑖 and 𝑗.
To help us, we find 𝐵𝐶 in terms
of 𝑖 and 𝑗. 𝐵𝐶 has a magnitude of 47 and is
pointing in the opposite direction to the unit vector 𝑗. And so it is negative 47𝑗. Back to 𝐴𝐶 now, looking at the
diagram, we can see that 𝐴𝐶 is 𝐴𝐵 plus 𝐵𝐶. And because we have 𝐴𝐵 and 𝐵𝐶
in terms of 𝑖 and 𝑗, it’s now simple to write 𝐴𝐶 in terms of 𝑖 and 𝑗. Let’s 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 𝑖 has components one, zero. Our vector 𝐴𝐵 which is 47𝑖 has
components 47, zero. And it’s a similar story for the
vector 𝐴𝐶. Here, we’ve also used the fact that
the unit vector 𝑗 has components zero, one.
Now we’re ready to find their dot
product. It’s the product of the
𝑥-components, so 47 times 47, plus the product of the 𝑦-components, that’s zero
times negative 47. Here, we’re using the general rule
for the dot product of vectors in component form. Of course, zero times negative 47
is just zero. And so we’re left with 47 times 47,
as before. So the dot product of the vectors
𝐴𝐵 and 𝐴𝐶 is 2209.
Here, we’ve 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, it’s 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.
