The Dot Product Level 4 In this video we are going to go over slightly
more challenging examples that make use of the dot product. Let’s jump straight into
the first example. If u is a unit vector find u dot v and u dot
w. Ok here we have an equilateral triangle with
vectors as sides, we are asked to find 2 separate dot products in this case u dot v and u dot
w. Let’s first find u dot v, according to the problem u is a unit vector this means
that it has a magnitude equal to one. Now this also means that vector v and vector w
also have a magnitude equal to one since they form the sides of an equilateral triangle
so all the sides have the same length. Recall from elementary geometry that an equilateral
triangle has 3 congruent angles each measuring 60 degrees so the angle between these vectors
is going to be equal to 60 degrees. Alright with this information we can use the
geometric definition of the dot product to find u dot v, substituting the magnitudes
and angle we obtain the following expression, simplifying we obtain one half as the final
answer. Next let’s find u dot w. In order to find
the dot product of these two vectors we need to align them tail to tail. So let’s go
ahead and move the tail of vector u and align it with the tail of vector w. Next, we need
to calculate the angle between these two vectors; we can easily find the angle by using the
linear pairs and subtracting 60 degrees from 180 degrees. In this case the angle between
them would be equal to 120 degrees. Now we go ahead and find the dot product, substituting
these values we obtain the following expression, simplifying the expression we obtain negative
one half as our final answer. Let’s try the next problem.
A street vendor sells a apples , b burritos, and c churros on a given day. The vendor charges
50 cents for the apples, 5 dollars for the burrito and 1 dollar for the churro. If vector
F equals a,b,c and vector P equals 0.5, 5, and 1 what is the meaning of the dot product F
dot P?
Ok this word problem is essentially asking us to interpret the dot product. Notice that
the components of vector F represent the number of apples, burritos and churros sold on a
given day. The components of vector P on the other hand represent the price of an apple,
burrito and churro respectively. Let’s compute the dot product with these vectors and analyze
the result. Using the component definition of the dot product we go ahead and multiply
each of the vectors components and add them together, doing that we obtain the following
expression, notice that this expression represents the vendors revenue for the day since you
are taking the number of each food item sold and multiplying it by its respective price.
So this expression is nothing more than the vendor’s total revenue for the day. Let’s
try the next problem. Find a vector perpendicular to vector v.
Here we have a planar vector and we are asked to find a vector that is perpendicular to
this vector. Recall that two vectors are perpendicular if the dot product is equal to 0. So let’s
go ahead and use this fact to find a vector that is orthogonal to vector v. Let’s assume
that the vector perpendicular to vector v is equal to vector P with x and y components
equal to a and b respectively. We want the dot product between these two vectors to be
equal to 0, so let’s go ahead and compute the dot product between these two vectors
and set the result equal to 0.
Doing that we obtain the following expression. Now we need to figure out the value of a and
b that would satisfy this equation. A trivial answer would be setting both a and b equal
to 0 this would be the case when the second vector represents the zero vector, recall
that this vector is perpendicular or orthogonal to any vector v. We need to find a nontrivial
solution. We can think about the classical algebra 1 problems where you were asked to
find a line perpendicular to another line, recall that the slopes of perpendicular lines
were negative reciprocals of one another in other words you switch the rise with the run
and flip the sign on the numerator or denominator. This problem is essentially the same problem
but in this case we are using vectors. So the value of a and b that solves this equation
would be 5 and negative 2 or negative 5 and 2 either values solves the equation. So a
vector that is perpendicular to vector v would be the vector with components 5, negative
2 or negative 5 and 2. In general if you have a planar vector with
components a and b, the vector with components b, negative a or negative b and a, will be
orthogonal to the first vector. Let’s try a slightly more challenging problem.
Find k so that vector A and vector B are perpendicular. Ok here we have two planar vectors A and B,
we are asked to determine the value of k that would make these two vectors perpendicular
or orthogonal, we can determine the value for k by first finding an expression for the
dot product and then setting it equal to zero. Let’s first find an expression for the dot
product by using the component definition.
Doing that we obtain the following expression, then we set this expression equal to 0. Now
we are face to face with a linear equation. So we go ahead and solve for k which ends
up being equal to three halves. So by replacing k with three halves both vectors will be perpendicular
to one another. Alright let’s go over the final example.
For what values of b are the vectors -6, b, 2 and b, b squared, b orthogonal?
Ok this problem is very similar to the previous example we need to determine the values of
b that makes these two vectors orthogonal to one another. We once gain need to use the
component definition of the dot product and find an expression in terms of b and set it
equal to 0. So let’s go ahead and find an expression for the dot product, multiplying
the components and collecting like terms we obtain the following. We are now face to face
with a cubic polynomial, so we go ahead and factor out the common variable b from both
terms, and then we go ahead and rewrite the difference of squares into the factor b plus
2 and the factor b minus 2.
Finally it is just a matter of using the zero product property to solve for the values of
b. In this case we would end up with 3 solutions, 0, 2 and negative 2. By replacing the values
of b with 0, 2 or negative 2 we would generate vectors that are orthogonal to one another.
Notice that the zero vector is one of the solutions.
Ok in our next video we will go over how to find the angle between two vectors.