Now we'll look at an example
of subtraction of scalar multiples of two vectors.
We're given vectors c and d in component form
but we want to find negative three times vector c
minus two times vector d.
So we'll first perform the scalar multiplication,
meaning we'll multiply the components
of vector c by negative three.
Then we'll multiply the components
of vector d by positive two,
and then we'll find the difference
by subtracting the x and y components of the two vectors.
So we have negative three times vector c
minus two times vector d.
So we can write this as
"Negative three times vector c,"
in component form,
"Minus two times vector d," in component form.
And now we'll perform the scalar multiplication.
So we'd have an x component of
negative three times negative two, that's positive six.
And a y component of
negative three times negative three, that's nine.
Now we're going to keep this a subtraction problem.
So we're performing scalar multiplication
with a positive two.
So two times three is six, and two times two is four.
Just like arithmatic, we could write the subtraction problem
as an addition problem,
which we'll take a look at in a moment.
But now to find the difference,
we'll subtract the x components
and subtract the y components.
So we would have six minus six,
and nine minus four.
So the x component would be zero
and the y component would be five
of negative three times vector c
minus two times vector d.
This would be our resultant vector.
But I also want to show this on the coordinant plane,
but before we do that, it will be helpful to recognize
that negative three times vector c
minus two times vector d
is equivalent to negative three times vector c
plus negative two
times vector d.
Let's take a look at this on the coordinant plane.
Looking at the graph of vector c graphed here in blue,
negative three times vector c would be this vector here
in the opposite direction, but three times as long,
or having three times the magnitude.
And given vector d here, negative two times vector d
is the vector in the opposite direction
with twice the length or twice the magnitude graphed here.
And now that we know that negative three times vector c
minus two times vector d is equivalent to
negative three times vector c
plus negative two times vector d,
we want to find the sum of these two vectors.
On the coordinant plane there is two main ways to show this.
One way would be to take the vector negative two d
and place the initial point at the terminal point
of vector negative three c.
Which would look like this.
And the resultant vector would be this vector here,
at the initial point of negative three c
and the terminal point, at the terminal point
of negative two times vector d.
The second way to show the sum
would be to form a parallelogram
using the vector negative three c
and the vector negative two d.
Notice to form the parallelogram we would copy
vector negative three c here.
And the resultant vector would be the vector
that's the diagonal of the parallelogram
with the initial point at the origin.
Again we have the vector here
with an x component of zero and a y component of five.
I hope you found this helpful.