Section 11.3 The Dot Product
The previous section introduced vectors and described how to add them together and how to multiply them by scalars. This section introduces a multiplication on vectors called the dot product.
Note how this product of vectors returns a scalar, not another vector. We practice evaluating a dot product in the following example, then we will discuss why this product is useful.
Example 11.3.3. Evaluating dot products.
Solution 2. Video solution
The dot product, as shown by the preceding example, is very simple to evaluate. It is only the sum of products. While the definition gives no hint as to why we would care about this operation, there is an amazing connection between the dot product and angles formed by the vectors. Before stating this connection, we give a theorem stating some of the properties of the dot product.
Theorem 11.3.4. Properties of the Dot Product.
The last statement of the theorem makes a handy connection between the magnitude of a vector and the dot product with itself. Our definition and theorem give properties of the dot product, but we are still likely wondering “What does the dot product mean?” It is helpful to understand that the dot product of a vector with itself is connected to its magnitude.
The next theorem extends this understanding by connecting the dot product to magnitudes and angles. Given vectors and in the plane, an angle is clearly formed when and are drawn with the same initial point as illustrated in Figure 11.3.6.(a). (We always take to be the angle in as two angles are actually created.)
The same is also true of 2 vectors in space: given and in with the same initial point, there is a plane that contains both and (When and are co-linear, there are infinitely many planes that contain both vectors.) In that plane, we can again find an angle between them (and again, ). This is illustrated in Figure 11.3.6.(b).
Theorem 11.3.7. The Dot Product and Angles.
Note how on the left hand side of the equation, we are computing the dot product of two unit vectors. Recalling that unit vectors essentially only provide direction information, we can informally restate Theorem 11.3.7 as saying “The dot product of two directions gives the cosine of the angle between them.”
When is an acute angle (i.e., ), is positive; when when is an obtuse angle ( ), is negative. Thus the sign of the dot product gives a general indication of the angle between the vectors, illustrated in Figure 11.3.9.
Each image has the two vectors and along with shown. The dot product along with three possible cases of <, > and equal to are shown. In the first image, the dot product has a positive value, the angle between the two vectors is acute. In the second image, the dot product is equal to the angle is degrees. In the third image, the dot product has a negative value and the angle is obtuse.
We can use Theorem 11.3.7 to compute the dot product, but generally this theorem is used to find the angle between known vectors (since the dot product is generally easy to compute). To this end, we rewrite the theorem’s equation as
We practice using this theorem in the following example.
Example 11.3.10. Using the dot product to find angles.
The starts at the origin and ends at point The also starts at the origin and ends at point and is also drawn from the origin and ends at point
Solution 1.
Solution 2. Video solution
We see from our computation that as indicated by Figure 11.3.11. While we knew this should be the case, it is nice to see that this non-intuitive formula indeed returns the results we expected.
We do a similar example next in the context of vectors in space.
Example 11.3.12. Using the dot product to find angles.
All three angles between these vectors was or We know from geometry and everyday life that angles are “nice” for a variety of reasons, so it should seem significant that these angles are all Notice the common feature in each calculation (and also the calculation of in Example 11.3.10): the dot products of each pair of angles was 0. We use this as a basis for a definition of the term orthogonal, which is essentially synonymous to perpendicular.
Definition 11.3.14. Orthogonal.
Example 11.3.15. Finding orthogonal vectors.
Solution 1.
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Recall that a line perpendicular to a line with slope
has slope the “opposite reciprocal slope.” We can think of the slope of as its “rise over run.” A vector orthogonal to will have slope There are many such choices, though all parallel: -
There are infinitely many directions in space orthogonal to any given direction, so there are an infinite number of non-parallel vectors orthogonal to
Since there are so many, we have great leeway in finding some. One way is to arbitrarily pick values for the first two components, leaving the third unknown. For instance, let If is to be orthogonal to then soSo is orthogonal to We can apply a similar technique by leaving the first or second component unknown. Another method of finding a vector orthogonal to mirrors what we did in part 1. Let Here we switched the first two components of changing the sign of one of them (similar to the “opposite reciprocal” concept before). Letting the third component be 0 effectively ignores the third component of and it is easy to see thatClearly and are not parallel.
Solution 2. Video solution
An important construction is illustrated in Figure 11.3.16, where vectors and are sketched. In Figure 11.3.16.(a), a dotted line is drawn from the tip of to the line containing where the dotted line is orthogonal to In Figure 11.3.16.(b), the dotted line is replaced with the vector and is formed, parallel to It is clear by the diagram that What is important about this construction is this: is decomposed as the sum of two vectors, one of which is parallel to and one that is perpendicular to It is hard to overstate the importance of this construction (as we’ll see in upcoming examples).
The vectors and as shown in Figure 11.3.16.(b) form a right triangle, where the angle between and is labeled We can find in terms of and
Two vectors and are shown that start from the same position, the angle between them is marked the vector is shorter. From the end of a dashed perpendicular is drawn on
Two vectors and are shown that start from the same position, the angle between them is marked the vector is shorter. From the end of a dashed perpendicular is drawn on The vector and are also added. From the initial point the vector is drawn in the same direction as but it ends at the start of the perpendicular. The perpendicular starts at the tip of and ends at the tip of and is labeled
We also know that is parallel to to that is, the direction of is the direction of described by the unit vector The vector is the vector in the direction with magnitude
Since this construction is so important, it is given a special name.
Definition 11.3.17. Orthogonal Projection.
Example 11.3.19. Computing the orthogonal projection.
Solution 1.
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Applying Definition 11.3.17, we haveVectors
and are sketched in Figure 11.3.20. Note how the projection is parallel to that is, it lies on the same line through the origin as although it points in the opposite direction. That is because the angle between and is obtuse (i.e., greater than ).The axis is drawn from to and the axis is drawn from to Two vectors and are shown, both start at the origin. The vector ends in point and lies in the second quadrant and ends in point and lies in the first quadrant. The projection of on is shown and it lies in the third quadrant.Figure 11.3.20. Sketching the three vectors in Part 1 of Example 11.3.19 -
Apply the definition:These vectors are sketched in Figure 11.3.21.(a), and again in Figure 11.3.21.(b) from a different perspective. Because of the nature of graphing these vectors, the sketch in Figure 11.3.21.(a) makes it difficult to recognize that the drawn projection has the geometric properties it should. The graph shown in Figure 11.3.21.(b) illustrates these properties better.
Figure 11.3.21. Sketching the three vectors in Part 2 of Example 11.3.19
Solution 2. Video solution
We can use the properties of the dot product found in Theorem 11.3.4 to rearrange the formula found in Definition 11.3.17:
The above formula shows that the orthogonal projection of onto is only concerned with the direction of as both instances of in the formula come in the form the unit vector in the direction of
A special case of orthogonal projection occurs when is a unit vector. In this situation, the formula for the orthogonal projection of a vector onto reduces to just as
This gives us a new understanding of the dot product. When is a unit vector, essentially providing only direction information, the dot product of and gives “how much of is in the direction of ” This use of the dot product will be very useful in future sections.
Two vectors and are shown that start from the same position, the angle between them is marked the vector is shorter. From the end of a dashed perpendicular is drawn on The projection of and are also added. The projection of is in the same direction as but it ends at the start of the perpendicular. The perpendicular starts at the tip of the projection of and ends at the tip of and is labeled
Now consider Figure 11.3.22 where the concept of the orthogonal projection is again illustrated. It is clear that
This is not nonsense, as pointed out in the following Key Idea. (Notation note: the expression “ ” means “is parallel to ” We can use this notation to state “ ” which means “ is parallel to ” The expression “ ” means “is orthogonal to ” and is used similarly.)
Key Idea 11.3.23. Orthogonal Decomposition of Vectors.
Let nonzero vectors and be given. Then can be written as the sum of two vectors, one of which is parallel to and one of which is orthogonal to
We illustrate the use of this equality in the following example.
Example 11.3.24. Orthogonal decomposition of vectors.
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Let
and as in Example 11.3.19. Decompose as the sum of a vector parallel to and a vector orthogonal to -
Let
and as in Example 11.3.19. Decompose as the sum of a vector parallel to and a vector orthogonal to
Solution 1.
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In Example 11.3.19, we found that
LetIs orthogonal to (i.e., is ?) We check for orthogonality with the dot product:Since the dot product is 0, we know Thus: -
We found in Example 11.3.19 that
Applying the Key Idea, we have:We check to see ifSince the dot product is 0, we know the two vectors are orthogonal. We now write as the sum of two vectors, one parallel and one orthogonal to
Solution 2. Video solution
We give an example of where this decomposition is useful.
Example 11.3.25. Orthogonally decomposing a force vector.
Consider Figure 11.3.26.(a), showing a box weighing 50lb on a ramp that rises 5ft over a span of 20ft. Find the components of force, and their magnitudes, acting on the box (as sketched in Figure 11.3.26.(b)):
Image of a box being placed on an inclined plane. The downward force for gravity is marked as The surface of the ramp is labeled as base of the ramp is labeled and the height is marked as
Solution 1.
As the ramp rises 5ft over a horizontal distance of 20ft, we can represent the direction of the ramp with the vector Gravity pulls down with a force of 50lb, which we represent with
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To find the force of gravity in the direction of the ramp, we computeThe magnitude of
is Though the box weighs 50lb, a force of about 12lb is enough to keep the box from sliding down the ramp. -
To find the component
of gravity orthogonal to the ramp, we use Key Idea 11.3.23.The magnitude of this force is lb. In physics and engineering, knowing this force is important when computing things like static frictional force. (For instance, we could easily compute if the static frictional force alone was enough to keep the box from sliding down the ramp.)
Solution 2. Video solution
Subsection 11.3.1 Application to Work
In physics, the application of a force to move an object in a straight line a distance produces work; the amount of work is (where is in the direction of travel). The orthogonal projection allows us to compute work when the force is not in the direction of travel.
A box is shown that is being pushed to the right by a force. The displacement is shown as a horizontal vector and is labeled Force is drawn from the middle of the box at an angle and is labeled as The projection of the is also shown, it is parallel to the displacement vector.
Consider Figure 11.3.27, where a force is being applied to an object moving in the direction of (The distance the object travels is the magnitude of ) The work done is the amount of force in the direction of times
The expression will be positive if the angle between and is acute; when the angle is obtuse (hence is negative), the force is causing motion in the opposite direction of resulting in “negative work.” We want to capture this sign, so we drop the absolute value and find that
Definition 11.3.28. Work.
Let be a constant force that moves an object in a straight line from point to point Let The work done by along is
Example 11.3.29. Computing work.
A man slides a box along a ramp that rises 3ft over a distance of 15ft by applying 50lb of force as shown in Figure 11.3.30. Compute the work done.
Image of a box being placed on an inclined plane or ramp. The base of the ramp is labeled and the height is marked as Force is drawn from the middle of the box at an angle of from the horizontal and is labeled as
Solution.
The figure indicates that the force applied makes a angle with the horizontal, so The ramp is represented by The work done is simply
Note how we did not actually compute the distance the object traveled, nor the magnitude of the force in the direction of travel; this is all inherently computed by the dot product!
The dot product is a powerful way of evaluating computations that depend on angles without actually using angles. The next section explores another “product” on vectors, the cross product. Once again, angles play an important role, though in a much different way.
Exercises 11.3.2 Exercises
Terms and Concepts
1.
2.
How are the concepts of the dot product and vector magnitude related?
3.
How can one quickly tell if the angle between two vectors is acute or obtuse?
4.
Give a synonym for “orthogonal.”
Problems
Exercise Group.
In the following exercises, find the dot product of the given vectors.
11.
12.
Exercise Group.
In the following exercises, find the measure of the angle between the two vectors in radians.
Exercise Group.
17.
Find two nonzero vectors orthogonal to
18.
Find two nonzero vectors orthogonal to
19.
Find two nonzero vectors orthogonal to
20.
Find two nonzero vectors orthogonal to
Exercise Group.
In the following exercises, vectors and are given. Find the orthogonal projection of onto and sketch all three vectors with the same initial point.
Exercise Group.
In the following exercises, vectors and are given. Write as the sum of two vectors, one of which is parallel to (or is zero) and one of which is orthogonal to Note: these are the same pairs of vectors as found in Exercises 21–26.
27.
28.
29.
30.
31.
32.
33.
A 10lb box sits on a ramp that rises 4ft over a distance of 20ft. How much force is required to keep the box from sliding down the ramp?
34.
A 10lb box sits on a 15ft ramp that makes a angle with the horizontal. How much force is required to keep the box from sliding down the ramp?
35.
How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of to the horizontal?
36.
How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of to the horizontal?
37.
How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, with a force of 50lb applied horizontally?
38.
How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, with a force of 50lb applied at an angle of to the horizontal?
39.
How much work is performed in moving a box up the length of a 10ft ramp that makes a angle with the horizontal, with 50lb of force applied in the direction of the ramp?
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