The Dot Product Level 3

In this video we will go over a couple of examples that make use of both the geometric

definition and component definition of the dot product. Let’s go ahead and jump straight

into the first example. Let u, v and w be vectors in a plane or in

space. Which of the following expressions are meaningful? Which are meaningless? Explain

your answer. We need to determine if the expression

shown has any meaning. We can determine this by remembering the end result of scalar multiplication

and the scalar product, recall that one produces a vector and the other produces a scalar.

Let’s take a look at the first expression, the quantity v dot u dotted with w. Ok let’s break this expression into pieces

and analyze each operation. V dot u is going to produce a scalar which will then be dotted

with vector w. Recall that the dot product is defined only for vectors. Here we have

a scalar dotted with a vector so this expression is meaningless mainly because a scalar dotted

with a vector is not defined. The next expression is the quantity v dot

u times vector w. The first operation v dot u will simplify to a scalar which will then

be multiplied by vector w, so this expression is nothing more than scalar multiplication

so this expression is meaningful. Notice that the only difference between this expression

and first expression is that in this example we are multiplying and not taking the dot

product like we did in the first example. The next expression is the magnitude of vector

v times the quantity u dot w. Ok the quantity u dot w will simplify to a scalar which will

then be multiplied by the magnitude of vector v which is also a scalar. This is essentially

ordinary multiplication of real numbers so this expression is meaningful. The next expression is vector v dotted with

the quantity of vector u plus vector w. The sum of vector u and vector w will reduce to

a vector which will then be dotted with vector v. Since the dot product is defined only for

2 vectors this expression is meaningful and will reduce to a scalar. Up next is vector v dot u plus vector w. The

first operation v dot u will reduce to a scalar which will then be added to vector w. Recall

that it is not possible to add a scalar with a vector so this expression is meaningless. Finally the magnitude of vector v dotted with

the quantity of vector u plus vector w. The sum of vector u and vector w will simplify

to a vector which will then be dotted with the magnitude of vector v (a scalar) since

you cannot dot a scalar with a vector this expression is meaningless. Alright let’s

try a different type of problems. Find the dot product of two vectors if their

lengths are 6 and 1/3 and the angle between them is pi over 4.

In this problem we are given the length or magnitude of two vectors in this case 6 and

1/3. We are also given the angle between the vectors in radians. We have at our disposal

two separate expressions to compute the dot product, the geometric definition and the

component definition, for this example we need to use the first definition since we

have the magnitudes and the angle between the vectors. Substituting the values into

the geometric definition of the dot product we obtain the following, then we simplify

the expression, lastly we go ahead and rationalize the denominator obtaining the square root

of 2 as the final answer. Alright let’s move along to the next example.

Given vector a and vector b, find a dot b. In this example we have 2 planar vectors in

component form this means that we can use the component definition of the dot product

to find a dot b. So we go ahead and multiply the x-components of each vector together and

add this result to the product of each vectors y-components. The first product simplifies

to 12 and second product simplifies to negative 6, then it is just a matter of adding these

numbers together obtaining 6 as the final answer. Notice that we can find the dot product

in two distinct ways it all depends on what parts of the vectors are provided to you.

Alright let’s try the next one. Here we have two vectors in space since we

have 3 components. The dot product can be computed similarly to the previous example.

The only difference is that we have 3 components. So we go ahead and use the component definition

of the dot product. We first need to multiply the x-components of each vector together and

add this value to the product of the vectors y-components which in turn are added to the

product of the vectors z-components. Simplifying the first product we obtain 15, the second

product reduces to 0 and the final product simplifies to negative 20. In the end the

dot product is equal to negative 5. Let’s try the next example.

Here we have two vectors written in unit vector form recall that the unit vectors i, j and

k represent the x, y and z components respectively. So we can easily find the dot product by just

using the component definition of the dot product, in this case we will multiply the

coefficients of i hat together and then add this value to the product of the coefficients

of j hat, and then add this value to the product of the coefficients of k hat. The first product

simplifies to 5, the second product reduces to zero and the third product simplifies to

27. Adding the expressions together we obtain the final answer equal to 32. Alright let’s

try a slightly more challenging example. Ok here we have vector A with components equal

to s, 2s, and 3s, and vector B with components t, negative t and 5t. This might seem a little

bit confusing since we do not have real numbers representing the components; in this case

they have been replaced by variable expressions. All this means is that the dot product is

going to simplify to a variable expression as oppose to a scalar. So we go ahead and

use the component definition of the dot product. We approach the problem in the same way as

we did in the previous examples. So we multiply the x, y and z-components together and add

them. The first product reduces to st, the second product reduces to negative 2st and

third product reduces to 15st, finally we collect like terms and simplify obtaining

14st as the final answer. Ok in our next video we will go over slightly more challenging

examples.