We're given vector A and B,
and asked to find the dot product of vector A and B.
To find the dot product of two vectors,
we find the sum of the products
of the corresponding components.
The value of the dot product does give us
information about the angle theta between the two vectors,
which we will discuss more about after our first example.
So, to find vector A dotted with vector B,
let's write vector A and component form.
We would have an X component of negative three,
a Y component of positive five,
we're dotting this with vector B in component form,
which would be four coma negative two.
The dot product would be equal to negative three
times positive for, plus five times negative two.
Again, that would be negative three times positive four,
plus five times negative two,
which is equal to negative 12 plus negative 10,
which is equal to negative 22.
Notice in this case, our dot product
is equal to negative 22.
Depending on whether the dot product is positive,
negative, or zero, it does give us information
about the angle theta between the two vectors.
When the dot product is equal to zero,
the two vectors are orthogonal, or perpendicular,
and theta is equal to pi over two radians.
If the dot product is positive, or greater than zero,
then theta is less than pi over two radians,
and if the dot product is less then zero, which is our case,
meaning the dot product is negative,
then the angle theta it would be
greater than pi over two radians.
If we graph these two vectors,
the angle between the two vectors should be
greater than pi over two radians.
Let's verify that.
Vector A is graphed here in blue,
and vector B is graphed here in red.
The angle theta between the two vectors
would be this angle here.
Notice how this angle is greater than pi over two radians.
We have an obtuse angle between vector A and vector B.
Let's take a look at a second example.
Same question just different vectors.
We have vector A, component form, which would be
one coma negative six dotted with vector B
in component form, which would be three coma negative two.
The dot product would be equal to one times three,
plus negative six times negative two.
Again, one times three, plus
negative six times negative two,
which would be three, this would be plus 12,
which is equal to positive 15.
Because the dot product, in this case, is positive,
the angle theta between these two vectors
is less than pi over two radians.
Again, let's go ahead and verify that.
Vector A is graphed in blue,
and vector B is graphed in red.
The angle theta between these two vectors
is this angle here, and notice how
this angle is less than pi over two radians.
Here we have an acute angle between vector A and vector B.
Let's take a look at one more example.
We have vector A, with an X component of two
and a Y component of negative four,
dotted with vector B, which has an X component
of negative four, and a Y component of negative two.
The dot product will be equal to two times negative four,
plus negative four times negative two.
Again, we have two times negative four,
plus negative four times negative two,
which is equal to negative eight, here we have plus eight,
so here, we have a dot product which is equal to zero,
and therefore, the two vectors are orthogonal,
or perpendicular, and theta is equal to pi over two radians.
Looking at the graph of the two vectors,
notice we have vector A in blue, vector B in red.
The angle theta between the two vectors, in this case,
is a right angle, verifying that these two vectors
are orthogonal, or perpendicular.
I hope you found these examples helpful.