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After watching this lesson, you will be able to explain what a dot product is, how it is different from a cross product, and use equations to calculate dot products. A short quiz will follow.

What Is a Dot Product?

A vector is a quantity that has both magnitude (numerical size) AND direction. A scalar, on the other hand, is a quantity that has only magnitude - it's just a number. Vectors are things like velocity, displacement, force, or electric field. Those quantities always have a direction. A velocity isn't 3 m/s; it's 3 m/s NORTH. But a scalar would be something like temperature, time or distance. Those are defined by their numerical values - they're not pointed in any particular direction.

When we multiply two numbers together that happen to be vectors, we do it in different ways depending on the circumstances. If the result you'll be getting is a vector quantity - if it has a direction itself - then you have to do a cross product. But if the result you want is a scalar quantity, a quantity that doesn't have a direction, then you use a dot product. A dot product is where you multiply one vector by the component of the second vector, which acts in the direction of the first vector.

So, for example, work is force multiplied by displacement. It's two vectors multiplied together. But more specifically it's the force acting in the direction you're moving, multiplied by the displacement. This is why work is a dot product. Of course, the fact that work is a scalar quantity, not a vector, might also have tipped you off.

Equations for Dot Product

There are two equations for dot products. One for when you have the overall magnitudes and angles of the vector, which looks like this:

Dot product equation: overall magnitudes and angles

And another, if you're given the two vectors in component form, which looks like this:

Dot product equation: component form

So, for the first, if you're multiplying vector A and vector B together, you take the magnitude of vector A, multiply it by the magnitude of vector B, and multiply that by the cosine of the angle between them.

But for the second equation, a situation where you don't know the overall magnitudes, you multiply the x components and y components together separately and then add them up. The answer works out exactly the same.

So, with our example of work, if you know that the force is 3 newtons and the displacement is 2 meters and the angle between the force and displacement is 30 degrees, you multiply 3 by 2 by cosine 30 and you get your answer. But, if instead you're told that the force is 30i + 8j newtons, and the displacement is 15i + 15j meters, you would instead use the second equation. It's all based on what information you're given.

Example Calculations

Alright, let's go through an example. Imagine you're sweeping the floor with a broom. You're pushing down on the broom with a force of 50 newtons, and the broom is at a 50 degree angle with the floor. So you're pushing partly down and partly forwards. If you push the broom 3 meters across the floor, how much work is done?

Well, work is a scalar and is equal to force multiplied by displacement, so we definitely want to do a dot product (I'm sure you're very surprised). First of all, let's write out what we know. The force, F, is 50 newtons and the displacement, x, is 3 meters.

Diagram for example

Looking at this diagram, we can see that the angle between the force and the displacement is the same as the angle between the broom and the floor - 50 degrees. So theta equals 50 degrees. Plug all that into our dot product equation, and we get 50 multiplied by 3 multiplied by cosine 50, and that gives us 96.4 Joules of work. And that's it; that's our answer.

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But what if we weren't given the 50 newtons or the 3 meters? What if instead we were given them in component form? We were told that we were pushing on the broom with a force of 32.1i - 38.3j newtons (or, in other words, 32.1 newtons forwards and 38.3 newtons down). And we were told that the displacement was 3i meters; there's no y-component here because the broom is moving directly horizontally.

Well, then we would have to use the other equation for the dot product. Multiply the x-components, 32.1 multiplied by 3, and multiply the y-components, -38.3 multiplied by zero, and we get 96.3 Joules of work. Basically, exactly the same number (it's only slightly different because of rounding).

So hopefully this example proves to you that the equations really do work!

Lesson Summary

A vector is a quantity that has both magnitude (numerical size) AND direction. A scalar, on the other hand, is a quantity that only has a magnitude - it's just a number. When we multiply two numbers together that happen to be vectors, we do it in different ways depending on the circumstances. If the result you'll be getting is a vector quantity - if it has a direction itself, then you have to do a cross product. But if the result you want is a scalar quantity, a quantity that doesn't have a direction, then you use a dot product. A dot product is where you multiply one vector by the component of the second vector, which acts in the direction of the first vector.

There are two equations for dot products - one for when you have the overall magnitudes and angles of the vector, which looks like this:

Dot product equation: overall magnitudes and angles

And another, if you're given the two vectors in component form, which looks like this:

Dot product equation: component form

In the first equation, the angle is the angle between the two vectors. In the second, you have to multiply the x components and the y components, and then add the totals up.

The most common and important example of a dot product is calculating work: force multiplied by displacement. This is because it's really force in the direction you're moving, multiplied by displacement. But any physics situation where you're multiplying two vectors and getting a result that's a scalar will be a dot product.

Learning Outcomes

After you have finished this lesson, you'll be able to:

Define vector and scalar

Explain what a dot product is

Identify the two equations for dot products

Describe how to calculate work using the two dot product equations

The dot product is a method to multiply two vectors that results in a scalar. Calculating work in physics requires the dot product. Let's work through some problems utilizing the dot product. Solve the problems on your own without a look at the solutions. After you are done, check your answers under the solutions section.

Practice Problems

1. Determine the dot product between vector a, which has a magnitude of 2 and vector b, which has a magnitude of 6. The angle between vectors a and b is 30°.

2. The dot product of two vectors is 3, vector p has a magnitude of 2, and vector q has a magnitude of 4. What is the angle between vectors p and q?

3. Vector a is 2 units east and 3 units north. Vector b is 4 units west and 2 units south. Determine the dot product of vectors a and b.

4. Determine the work done by a force of 10 N applied at a 90° angle to the displacement of the object, which is 20 m. Work is the dot product of the force on the object and displacement of the object.

Solutions

1. abcosθ = (2)(6)cos30° = 10.4

2.

pqcosθ = the dot product

(2)(4)cosθ = 3

8cosθ = 3

cosθ = 3/8

θ = 68°

3. ax bx + ay by = (2)(3) + (-4)(-2) = 6 + 8 = 14

4.

W = Fdcosθ

W = (10)(20)cos90°

W = 0 joules

Work can only be done on an object if a component or all of the force exerted on it is parallel or antiparallel to the object's displacement.

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