Hello!
Do you remember this situation from high school physics?
You were supposed to calculate "the work done" by a constant Force (F)
exerted on the car by moving it over a distance s.
At school you learned to take the product of the magnitudes of Force F,
distance s and the cosine of the angle between F and s.
This quantity is often written as F dot s and is called the dot product of Force and Displacement
if you regard Force and Displacement as vectors!
The big dot in the notation explains the name.
In this slide we used vector notation.
That's the reason that we wrote the magnitudes of the force and the displacement
of the previous slide as the lengths of the vectors F and s.
A disadvantage of calculating the dot product in this geometrical way is that you have to know
the angle between the vectors and the lengths.
Often however the components of the vectors are given and then the angle is hard to compute.
But If that's the case, I have good news:
then there is a much easier algebraic way to calculate the dot product!
Let's look how this works.
It is shown for two general three dimensional vectors a and b.
To calculate the dot product you just have to multiply the first components of the two vectors,
then the second ones and at last the third ones.
Adding the results gives the dot product.
With a little practice everyone can do it!
This is our formal definition of the dot product of two three dimensional vectors.
The dot product of two two dimensional vectors is defined in a similar fashion.
Here is a two-dimensional example:
Suppose the force F is the vector with components minus 4 and 3
and suppose you have moved the car over a distance of 3 meters.
Since the displacement is horizontal and in negative direction
you can describe this with a vector with components minus 3 and 0.
The work done is the dot product of the vectors F and s.
With our new formula you can simply calculate this dot product as the sum of the products
of the corresponding components.
In this example the answer is 12.
And now the three-dimensional case.
Suppose two vectors are given.
The dot product is then easily calculated,
again multiply the corresponding components
and add the results.
In this case the answer is minus 13.
In class I would let you think about the question that's on the slide now.
The dot product of a two and a three dimensional vector is asked for.
The answer is that it is not defined.
So remember:
only dot products of vectors with the same dimension can be calculated!
Before connecting the two different ways of calculating the dot product of two vectors,
one with a geometric approach and the other with an algebraic approach,
I will show you some nice properties to do meaningful calculations.
Let's have a look at these properties, valid for vectors of the same dimension.
The first states that the dot product is commutative,
a property well known from ordinary multiplication with numbers.
With numbers you are also used to the second property which combines the dot product with the sum,
better known as the distributive law.
The third property states the connection between dot product and scalar multiplication of vectors.
The fourth property, which is somewhat different, is easy to understand.
Think about it!
It states that the dot product of a vector and itself equals the square of its length.
A property that you are going to use often.
Especially the first three properties make you think that calculations with the dot product
are more or less the same as calculations with numbers.
But be careful!!
Some weird things can occur.
Curious?
Make the exercises!
One of these weird things is the fact that the dot product of three vectors isn't defined.
One reason for this is that the dot product of two vectors is a scalar.
The properties given here can all be checked by using the definition of the dot product.
Some of them are left to you as an exercise.
This example shows how you can use these properties to write the distance between two vectors
as a sum of dot products.
Useful since a geometrical approach is connected with an algebraic approach!
Using the fourth property you first write the square of the distance as a dot product.
Now using the first three properties you can rewrite the result as sum of dot products.
Now you of course want to know what the connection is between the geometric approach
this video is started with and the algebraic approach introduced a little later.
It appears to be a direct consequence of the well known Cosine Law which gives the connection
between the lengths of the three sides of a triangle and one of its angles.
Applying this law on the triangle formed by the common starting point and the two endpoints of two vectors,
you get the result as shown here in terms of lengths of vectors.
You can use the formula derived on the previous slide to rewrite the left hand side in terms of the dot product.
A little rewriting gives the connection searched for.
You may try to do it by yourself in the exercises.
All nice math,
but what else can be done with this concept!
Watch the next video to discover this!
But it's better to practice a little bit first.
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