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Dot Product of two Vectors in 3D:
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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Find
, where
, and
.
First, let's plot this on paper, using the usual 3-D axes with the x-axis out of the plane of the diagram. Draw the axes and then click "Next" to check your work.
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Now show the vector
A
, based at the origin. Then check your work by clicking "Next".
Note the use of color for the coordinates of the head of the vector.
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Now show the vector
B
, based at the origin. Then check your work by clicking "Next".
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How do we find the Dot Product?
We find the Dot Product from the sum of the products of the components.
Set that up in terms of the general components
of the vectors.
.
What is the value of
?
What is the value of
?
What is the value of
?
What are the values of the components of
B
?
Substitute the numerical values.
Simplify this.
We get
.
We got a positive result here from positive components. Let's try reversing both vectors so that they have negative components.
Start a new diagram and show
C
= –
A
.
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Show
D
= –
B
.
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Do we still use the sum of the products of the components to get the Dot Product?
Yes.
Even though the components are negative?
Yes.
Set up the Dot Product in terms of the general components
Should we still be using the plus sign in the middle of this expression?
Yes, we need the sum of the products.
Substitute the numerical values for the components.
Simplify.
We get
This result is the same as before, in spite of the negative components.
To see why, state the definition of the Dot Product in terms of the angle
theta
between the vectors.
,
where
C
and
D
are the lengths of the vectors.
Are the lengths always positive?
Yes.
What does determine the algebraic sign of the Dot Product?
The value of the cosine of
theta
.
What values of
theta
lead to a negative value of the cosine?
Values between 90 and 180 degrees.
So we see consistency between the acute angle between
C
and
D
, and the component method.
Now let's try an example with an obtuse angle, using the vectors
, and
.
Draw a diagram on paper showing vector
E
. Then check by clicking "Next".
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Add vector
F
. Then check by clicking "Next".
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Set up the Dot Product in terms of
Substitute the numerical values.
Simplify.
We get
.
This confirms that the Dot Product can be negative.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor
General Contents
Detailed Contents
Index