Lesson Video: Products of Vectors

In this video we learn how to multiply vectors using the dot product and cross product, as well as the physical phenomena these vector products describe.

12:51

Video Transcript

In this video, we’re going to learn about products of vectors, what they are, why they are useful, and how to calculate them. To give an idea of how useful products of vectors are in describing physical phenomena, imagine a situation where you’re moving a large heavy box across the floor. You push up on the box a bit such that the force vector points above the horizontal. Over time, you’re able to push the box along the floor such that it ends up with the horizontal displacement vector 𝑑. If we want to calculate how much work is done in moving the box that distance, we’ll need to make use of a product of vectors. Or imagine a different scenario. In this one, we’re sending a screw into a block of wood. We use a screwdriver to apply a force 𝐹, a displacement of π‘Ÿ away from the screw’s axis of rotation. By using a particular vector product, we’re able to tell which direction, whether into or out of the board, the screw goes.

As we start out, it can be helpful to contrast vector products with products we may be more familiar with. And that’s the products of scalars. Recall that a scalar is a quantity that has magnitude but no direction to it. Examples of scalars would be numbers such as five kilograms or 10 seconds or even 65 degrees. Each of these values has a magnitude without an associated direction. If we wanna find the product of two scalars, say five kilograms and 10 seconds. Then we simply multiply them as we’re used to multiplying scalar values. The result of this multiplication would be 50 kilogram-seconds. Vectors, of course, are quantities that have both magnitude and direction. Examples of vectors might be 10 meters to the east or two kilometers down.

Now if someone said what is the product of these two vectors, we might not be quite so confident as how to combine them together as we were with scalars. After all, how would we go about combining the different directions of east and down? It turns out that vector products come in two particular types. That let us navigate this tricky question of how to combine unlike things.

The first approach we have to finding the product of two vectors, we can call them 𝐴 and 𝐡, is to take what’s called the dot product. It looks like a dot placed between those two vectors. If we had vector 𝐴 and vector 𝐡 with their tails at the same location. When we take their dot product, what we’re doing is finding out how much of the vectors lie along one another. That’s the geometrical meaning of the dot product. And if we give the name πœƒ to the angle between these two vectors 𝐴 and 𝐡, we can write that the dot product of 𝐴 and 𝐡 is equal to the product of their magnitudes multiplied by the cos of that angle πœƒ. So the dot product is useful to us in finding the angle between vectors. With this mathematical relationship in front of us, we can try out a few simple cases to explore this equation.

Imagine we had two vectors, 𝐢 and 𝐷, that were perpendicular to one another. When we consider that the dot product means the overlap of two vectors β€” that is, how much of one lies along the other. By looking at this diagram, we might naturally guess that the dot product of 𝐢 and 𝐷 is zero since they’re perpendicular. This intuition is confirmed when we consider the equation for dot product. We know that the cos of 90 degrees, which is the angle between 𝐢 and 𝐷, is zero. So we’re correct in saying that this dot product itself is zero.

Going just a bit further, imagine that we are given 𝐴 and 𝐡 in their component forms, not graphically but numerically. And in this case, both vectors have three dimensions to them: 𝑖, 𝑗, and π‘˜. If we wanted to take the dot product of these two vectors, how would we do that mathematically? This gets a bit of the question we talked of earlier, of how do we combine or sort out unlike types as we have with different unit vectors in this example. The answer to this is that we multiply our vectors according to their type. 𝐴 π‘₯ times 𝐡 π‘₯ plus 𝐴 𝑦 times 𝐡 𝑦 plus 𝐴 𝑧 times 𝐡 𝑧, keeping the different vector components separate from one another. Then finally add those different component products together to find the dot product, which will ultimately be a scalar. Both these relationships are helpful to keep in mind when we’re working with dot products.

The second type of vector product is called a cross product. And it is symbolized with an x or a cross symbol in between our test vectors, 𝐴 and 𝐡. While the dot product creates a scalar, taking a cross product of two vectors creates a vector. Another difference with the dot product is that the dot product is commutative. Meaning that 𝐴 dot 𝐡 is equal to 𝐡 dot 𝐴. But the cross product is not like that. That is, in general, 𝐴 cross 𝐡 is not equal to 𝐡 cross 𝐴. So the order is very important when we calculate cross products. If we again had two vectors, vector 𝐴 and vector 𝐡, then we can understand the physical meaning of the cross product this way. When we cross 𝐴 into 𝐡, we are solving for the components that are perpendicular to the plane in which vector 𝐴 and vector 𝐡 lie. For example, if our two vectors 𝐴 and 𝐡 were in a plane with 𝑖- and 𝑗-components, then their cross product would have no components in that direction. It would be entirely either into the page or out of the page.

So when we think of cross products, think of results that are perpendicular to the input vectors we use. There is a very helpful mathematical tool for calculating cross products. The tool involves a three-by-three matrix of values, where we take the determinant of that matrix in order to solve for the cross product. The three common unit vectors 𝑖, 𝑗, and π‘˜ are the head of each column. And the two vectors we’re crossing together, 𝐴 and 𝐡, have their components listed as the last two rows of this matrix. If we were to calculate the determinant of this matrix, we’d find a result with an 𝑖-, 𝑗-, and π‘˜-component. In other words, our result would be a vector. The particular value for each of those three components depends on the values of 𝐴 and 𝐡. But notice as we look at the three different component results, the components of 𝐴 and 𝐡 that go into each one are perpendicular to that component direction.

For example, with 𝑖, the only components of 𝐴 and 𝐡 that affect 𝑖 are in the 𝑗- and π‘˜-directions. In a similar way, it’s the components in the 𝑖- and π‘˜- directions that determine the magnitude of the 𝑗-component. And it’s the components of 𝐴 and 𝐡 in the 𝑖- and 𝑗-direction that determine the magnitude of the π‘˜-component. This confirms what we spoke of earlier where the cross product has to do with results perpendicular to the direction of the input vectors. With all that as background, let’s try a couple of example problems having to do with vector products.

Find the angle between the two vectors 𝐲 equals 2.0𝐒 plus 4.0𝐣 plus 8.0𝐀 and 𝐳 equals 6.0𝐒 plus 4.0𝐣 plus 6.0𝐀.

If we call the angle between these two vectors πœƒ, it’s that we want to solve for. To start out, we can recall that the dot product between two vectors, 𝐀 and 𝐁, is equal to the product of their magnitudes times the cosine of the angle between them. And we can recall further that the dot product is also equal to the product of the π‘₯-components plus the product of the 𝑦-components plus the product of the 𝑧-components of our two vectors. Bringing those two relationships together, we can write that, in our case, the product of the π‘₯-components of our vectors plus the product of the 𝑦-components plus the product of the 𝑧-components. Is equal to the product of their magnitudes multiplied by the cos of πœƒ, the angle between them. If we divide both sides of this equation by the magnitude of 𝐲 times the magnitude of 𝐳 and then take the inverse cosine of both sides. So πœƒ, the angle we’re interested in, is equal to the inverse cos of 𝑦 π‘₯, 𝑧 π‘₯ plus 𝑦 𝑦, 𝑧 𝑦 plus 𝑦 𝑧, 𝑧 𝑧 divided by the product of the magnitudes of our two vectors.

We can recall that the magnitude of a vector, when it has three dimensions, is equal to the square root of the π‘₯-dimension squared plus the 𝑦-dimension squared plus the 𝑧-dimension squared. If we insert the magnitude expansion for both 𝑦 and 𝑧, then we now have an expression for πœƒ entirely in terms of the components of our two vectors. In the case of our vector 𝐲, 2.0 is 𝑦 π‘₯, 4.0 is 𝑦 𝑦, and 8.0 is 𝑦 𝑧. And for 𝐳, 6.0 is 𝑧 π‘₯, 4.0 is 𝑧 𝑦, and 6.0 is 𝑧 𝑧. When we plug each of these values in where it fits in our equation. With all these values plugged in, when we enter this expression on our calculator, we find πœƒ is 28 degrees. That’s the angle between our two vectors 𝐲 and 𝐳.

That’s an example using the dot product combination of vectors. Now let’s try an example using the cross product.

Calculate the cross product of 𝐲 equals two 𝐒 plus four 𝐣 plus eight 𝐀 and 𝐳 equals six 𝐒 plus four 𝐣 plus two 𝐀.

In this example, we want to solve for 𝐲 cross 𝐳, where 𝐲 and 𝐳 are both three-dimensional vectors. We can begin our solution by recalling the mathematical definition for a cross product. The cross product of two vectors, we can call them 𝐀 and 𝐁, is equal to the determinant of a three-by-three matrix , where the columns are headed by the three unit vectors 𝐒, 𝐣, and 𝐀. And the last two rows are populated by the respective components of vectors 𝐀 and 𝐁. We can apply this formula to our vectors 𝐲 and 𝐳. As we set up our matrix, we know the top row will be the 𝐒, 𝐣, and 𝐀 unit vectors. The next row will be the respective components of the vector 𝐲. Looking at 𝐲, we see that it’s π‘₯-component is two, its 𝑦-component is four, and its 𝑧-component is eight. So we write those values in to our matrix. In the next row, we’ll write the components of 𝐳. The π‘₯-component of 𝐳 is six. Its 𝑦-component is four. And its 𝑧-component is two.

Now we’re ready to calculate the determinant of this matrix and solve for the cross product of 𝐲 cross 𝐳. When we compute this determinant, we find that the 𝐒-component is eight minus 32, or negative 24. The 𝐣-component is four minus 48, or negative 44. And the 𝐀-component is eight minus 24, which equals negative 16. Our overall cross product then is negative 24𝐒 minus 44𝐣 minus 16𝐀. That’s the cross product of the vectors 𝐲 and 𝐳.

To summarize, products of vectors are useful because they help us to understand and themselves reveal physical phenomena. These could include the direction to point a windmill or the best way to get power out of an engine. There are two types of products of vectors that we’ve learned: the dot product, or the scalar product, and the cross product, which produces a vector. And finally, when calculating vector products, matrices can be helpful. For example, the dot product of 𝐀 and 𝐁 can be expressed as a combination of two matrices. And the cross product of 𝐀 and 𝐁 can be expressed as the determinant of a three-by-three matrix.

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