Lesson Video: Cross Product in 3D Mathematics

In this video, we will learn how to find the cross product of two vectors in space and how to use it to find the area of geometric shapes.

16:58

Video Transcript

In this video, we’re talking about the cross product in 3D. This is a mathematical operation that we apply to two vectors in three-dimensional space. We’re going to learn how to do this, why it’s useful, as well as the geometric meaning behind the cross product.

As we get started, the first thing we’ll say about this operation is that it applies to two vectors that may have as many as three dimensions each. For example, imagine that we have two vectors 𝐀 and 𝐁, each with an 𝐒-, 𝐣-, and 𝐀-component. The three-dimensional cross product combines vectors like these, and what results is itself a vector. And not only that, this vector is perpendicular to the original too.

So if we took the cross product of vectors 𝐀 and 𝐁, and that’s represented by this symbol Γ— like this, then this operation would give us a third vector, we’ve called it 𝐂, that’s perpendicular to 𝐀 and 𝐁. So say that this here is vector 𝐀 and this is vector 𝐁. And even though both are three-dimensional, we can imagine viewing them from a perspective perpendicular to both vectors. Viewed this way, the vector 𝐂 resulting from 𝐀 cross 𝐁 would point out of the screen at us in this direction.

And as far as the magnitude of vector 𝐂 goes, if we were to make a parallelogram where two adjacent sides are our two input vectors 𝐀 and 𝐁, then the area of that parallelogram is equal to the magnitude of 𝐀 cross 𝐁. And it’s this area that tells us the magnitude of 𝐂. And that tells us one more thing about the three-dimensional cross product, that when our two input vectors define adjacent sides of a parallelogram, that shape’s area is equal to the magnitude of the cross product of those vectors. Carrying out a cross product in 3D allows us to compute a number of useful physical quantities, for example, torque or angular momentum.

Now that we’ve looked into what the cross product means, let’s see how we actually go about calculating it. The way we mathematically represent a three-dimensional cross product is using a three-by-three matrix. In the first row of this matrix, we have the three orthogonal unit vectors 𝐒, 𝐣, and 𝐀. In the next row, we have the corresponding components of the first vector listed in our cross product, in this case vector 𝐀. And in the last row, we have those same components of our second vector.

Notice then that when we take a cross product, the order of the vectors we input matters. In general, 𝐀 cross 𝐁 is not the same as 𝐁 cross 𝐀. Now, note that we’ve explicitly indicated taking the determinant of our three-by-three matrix. It’s common though to represent the same mathematical operation this way. And it means the same thing. When we calculate this cross product, we’ll end up with a three-dimensional vector. The way we compute this result is component by component.

If we first calculate the 𝐒-component, this involves multiplying that unit vector by the determinant of the two-by-two matrix that remains if we cross out the row and column that contain the 𝐒 hat vector. And we can see that that determinant is equal to 𝐀𝑦 times 𝐁𝑧 minus 𝐀𝑧 times 𝐁𝑦. Whatever that works out to you, that is the 𝐒-component of our resulting vector.

We next move on to calculating the 𝐣-component. This involves a similar process, but note that we precede this component with a negative sign. Finding this component’s magnitude involves calculating the determinant once again of a two-by-two matrix, this time with these four elements. We see that’s equal to 𝐀π‘₯ times 𝐁𝑧 minus 𝐀𝑧 times 𝐁π‘₯. And lastly, we calculate the 𝐀 hat component.

Once more, crossing out the row and column in which this element appears, we find this component as the determinant of this two-by-two matrix, 𝐀π‘₯ times 𝐁𝑦 minus 𝐀𝑦 times 𝐁π‘₯. And with that, we’ve calculated the cross product 𝐀 cross 𝐁.

And recall that earlier we claimed that this result is perpendicular to both of our input vectors. To see if that holds true, let’s try this operation out on a test case. Say that we have two vectors 𝐕 one and 𝐕 two, where 𝐕 one is the 𝐒 hat unit vector and 𝐕 two is the 𝐣 hat unit vector. If we take the cross product 𝐕 one crossed with 𝐕 two, then according to our formula, that’s the determinant of a three-by-three matrix, where the second row and the third row, which we’ll fill in, are the respective components of 𝐕 one and 𝐕 two.

Since 𝐕 one is simply the 𝐒 unit vector, its magnitude in that direction is one and its magnitude in the other two directions is zero. Similarly, for 𝐕 two, its magnitude in the 𝐒 hat direction is zero. Its magnitude in the 𝐣 hat direction is one and zero in the 𝐀 hat. Calculating the resulting 𝐒-component of this cross product, it’s equal to the determinant of this matrix here. We see that equals zero times zero, which is zero. And from that, we subtract zero times one, also zero. This tells us that the cross product has no 𝐒-component itself. If we then calculate the 𝐣-component, that’s equal to negative one times one times zero, which is zero, minus zero times zero. So this cross product also has no 𝐣-component to it.

Lastly, we look at the 𝐀-component, equal to the determinant of this two-by-two matrix. One multiplied by one is one. And from that, we subtract zero times zero. Finally, then, we have a nonzero vector component. This whole cross product is simply equal to the 𝐀 hat unit vector. And recalling that 𝐕 one equals the 𝐒 hat vector and 𝐕 two equals the 𝐣 hat vector, we see that the resultant really is perpendicular to both of these input vectors.

Let’s continue to get practice with a cross product in 3D by working through a few examples.

Let 𝐕 equal the 𝐒 hat unit vector and 𝐖 equal three 𝐒 plus two 𝐣 plus four 𝐀. Calculate 𝐕 cross 𝐖.

So here we have these two vectors 𝐕 and 𝐖, and one of them is three-dimensional. To calculate their cross product then, we’ll want to recall the mathematical rule for calculating a three-dimensional cross product. Given two vectors 𝐀 and 𝐁 with as many as three dimensions each, 𝐀 cross 𝐁 is equal to the determinant of this three-by-three matrix.

Notice that the first row is populated by the three orthogonal unit vectors and the second and third rows by the corresponding components of our two input vectors 𝐀 and 𝐁. Remembering this rule, we can now apply it to our two particular vectors 𝐕 and 𝐖. We’ll set up 𝐕 cross 𝐖 like this. And with the unit vectors at the top of each column, we know that the components of 𝐕 are one, zero, zero in these terms, while the 𝐖-vector has a magnitude of three in the 𝐒-direction, two in the 𝐣-direction, and four in the 𝐀-direction.

Now we’re ready to calculate this cross product, and we’ll do it component by component, starting with 𝐒. That component will equal the determinant of this two-by-two matrix. That’s equal to zero times four, which is zero, minus zero times two, which is also zero. So the 𝐒-component of our cross product is zero. And now let’s look at the 𝐣-component. Its value is equal to negative one times the determinant of this two-by-two matrix. That’s one times four or four minus zero times three, which is zero. The 𝐣-component then has a nonzero value of negative four. And then, lastly, the 𝐀-component equal to the determinant of this two-by-two matrix. That’s one times two or two minus zero times three, or zero. So here is our resulting vector. This is the cross product of 𝐕 and 𝐖.

We could leave our answer in this form, or we can write it in vector form. Doing that, we use this angled bracket notation and we just write down the magnitude of each component. This is 𝐕 cross 𝐖.

Now let’s look at an example where all of the vectors involved are three-dimensional.

If vector 𝐀 equals three, four, negative four; vector 𝐁 equals two, five, negative four; and vector 𝐂 equals negative four, negative four, two, find 𝐀 minus 𝐁 cross 𝐂 minus 𝐀.

Okay, so here we have these three vectors 𝐀, 𝐁, and 𝐂 written out by their components. So, for example, considering vector 𝐀, this is its 𝐒-component, this is its 𝐣-component, and this is its 𝐀-component. Our goal is to calculate this cross product, where the two vectors involved are differences between two of our 𝐀, 𝐁, and 𝐂 vectors.

What we can do here is give these differences their own names. We’ll call 𝐀 minus 𝐁 vector 𝐃, and 𝐂 minus 𝐀 we’ll call vector 𝐄. What we’ll eventually do then is cross 𝐃 and 𝐄. But before that, we’ll need to figure out what each of these vectors are. Starting with vector 𝐃, we know that that’s equal to vector 𝐀, which is three, four, negative four, minus vector 𝐁, which is two, five, negative four. Working component by component, three minus two is one, four minus five is negative one, and negative four minus negative four is zero. So that’s vector 𝐃.

And vector 𝐄 is equal to vector 𝐂 here minus vector 𝐀 here. Negative four minus three is negative seven, negative four minus four is negative eight, and then two minus negative four is positive six. So we now have the two vectors 𝐃 and 𝐄. And we’ll write them off to the side that will combine in our cross product. To do that, let’s recall this rule for the three-dimensional cross product of two vectors. We’ve called them 𝐀 and 𝐁 here. This cross product equals the determinant of a three-by-three matrix.

In the first row, we have our unit vectors, in the second the corresponding components of vector 𝐀, and in the third those of vector 𝐁. So when we set up the cross product of 𝐃 and 𝐄, we’ll once again have 𝐒, 𝐣, and 𝐀 unit vectors at the top of each column. And in the next row, we’ll write in the components of vector 𝐃. Those are one, negative one, and zero. Lastly, the components of vector 𝐄: negative seven, negative eight, positive six.

Computing this cross product is now a matter of calculating the determinant of this matrix. Beginning with the 𝐒-component, if we cross out the row and column that contain this element, then this component is equal to the determinant of the two-by-two matrix remaining. This is equal to negative one times six or negative six minus zero times negative eight, or zero. So the 𝐒-component of our resulting vector is negative six.

And then we move on to calculate the 𝐣-component. This has a value of negative one times one times six, or six, minus zero times negative seven, or zero. This comes out to negative six 𝐣. And lastly, for our 𝐀-component, this is equal to one times negative eight, or negative eight, minus negative one times negative seven, which comes out to negative seven. Our 𝐀-component then simplifies to negative 15.

And now we’ve computed all three components of a resulting cross product. The vector we end up with is negative six 𝐒 minus six 𝐣 minus 15𝐀. Going back to our original three vectors 𝐀, 𝐁, and 𝐂, this equals 𝐀 minus 𝐁 cross 𝐂 minus 𝐀.

Let’s now look at an example where we consider the geometric meaning of the cross product.

𝐴𝐡𝐢𝐷 is a parallelogram with vector 𝐀𝐁 equals negative one, one, three and vector 𝐀𝐃 equals three, four, one. Find the area of 𝐴𝐡𝐢𝐷. Give your answer to one decimal place.

Okay, so here 𝐴𝐡𝐢𝐷 is a parallelogram. And two of its sides are made of these two vectors we’re given, 𝐀𝐁 and 𝐀𝐃. So let’s say that this point here is point 𝐴. And we’ll say that it has the coordinates zero, zero, zero. In other words, 𝐴 is the origin in a three-dimensional set of axes.

From this point, we can draw vectors 𝐀𝐁 and 𝐀𝐃. If we say that they both lie in the plane of our screen, then vector 𝐀𝐁 could look like this and vector 𝐀𝐃 like this. We know that each of these two vectors connects two corners on our parallelogram. This tells us that point 𝐡 is right here at the end of vector 𝐀𝐁 and point 𝐷 is at the end of vector 𝐀𝐃. And because we’re working with a parallelogram, we can sketch in the remaining sides. They would look like this. And point 𝐢 in our parallelogram is here.

In our question, we’re asked to find this area, the area of 𝐴𝐡𝐢𝐷. To do this, we can recall the fact that the magnitude of the cross product of two vectors making up adjacent sides of a parallelogram is equal to that shape’s area. In other words, if we take vector 𝐀𝐁 and we cross it with vector 𝐀𝐃, then the magnitude of that cross product is equal to the area we’ve shaded in in blue.

Our first step then is to figure out how to calculate this cross product. We can recall that, given two three-dimensional vectors β€” we’ll call them 𝐀 and 𝐁 β€” 𝐀 cross 𝐁 is given by this expression. Here we have a three-by-three matrix where in the first row we have the unit vectors, in the second the corresponding components of 𝐀, and in the third those of 𝐁. So let’s now apply this general rule to the vectors given to us here, vectors 𝐀𝐁 and 𝐀𝐃.

Having started our three-by-three matrix, we’ll fill in the second row with the components of vector 𝐀𝐁. And we see that those are negative one, positive one, and three. And then the components of 𝐀𝐃 are positive three, positive four, and positive one. To calculate this determinant, let’s clear some space on screen and then calculate this cross product, starting with the 𝐒-component. We cross out the row and column containing this element. And the value of the 𝐒-component then is equal to the determinant of this two-by-two matrix. One times one is one. And from that, we subtract three times four, or 12.

Then, we move on to the 𝐣-component. This value equals negative negative one times one, or negative one, minus three times three, or nine. And lastly, there’s the 𝐀-component, equal to the determinant of this matrix. Negative one times four is negative four, minus one times three or three. These components simplify to negative 11𝐒, positive 10𝐣, and negative seven 𝐀. This then is the cross product of 𝐀𝐁 and 𝐀𝐃.

But we’re not quite to our answer because remember, we want to solve for the area of our parallelogram. To do that, we’ll need to calculate the magnitude of the vector we just computed. If we have a vector 𝐕 with components in the 𝐒-, 𝐣-, and 𝐀-directions, then we can recall that the magnitude of 𝐕 is equal to the square root of the sum of the squares of its components. We can apply the same rule to solve for the magnitude of 𝐀𝐁 cross 𝐀𝐃. Here, the components are negative 11, positive 10, and negative seven. If we square all these, we get 121, 100, and 49. And adding the squares together gives us 270. Entering the square root of 270 on our calculator, if we round it to one decimal place, the answer we get is 16.4. This then, in whatever relevant area units, is the area of 𝐴𝐡𝐢𝐷.

Let’s finish up now by reviewing some key points about the cross product in 3D. In this lesson, we learned that a cross product in 3D combines two three-dimensional vectors and yields another vector perpendicular to the original two. So if vectors 𝐀 and 𝐁 were lying in the plane of our screen, then 𝐀 cross 𝐁 would point out of the screen at us.

Along with this, we learned that the magnitude of a cross product gives the area of the parallelogram defined by the two vectors in the cross product. And lastly, we saw that if 𝐀 and 𝐁 are two three-dimensional vectors, then 𝐀 cross 𝐁 is equal to the determinant of this three-by-three matrix. Written in component form, that looks like this.

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