Learn about Calculating dot and cross products with unit vector notation

Khan Academy Presents: Calculating the dot and cross products when vectors are presented in their x, y, and z (or i,j, and k) components.

So far when I told you about the dot and the cross-products I give you the definition as the magnitude times the cosine or the sine of the angle between them. But what if you are not given the Vectors visually, what if you are not given the angle between them? How do you calculate the dot and the cross-products? So let’s say, let me give you the definition that I given you already. So let’s say I have (a.b), dot product that is the magnitude of a times the magnitude of b times cosine of the angle between them. a cross b is equal to the magnitude of (a) times the magnitude of (b) times sine of the angle between them. So kind of the perpendicular projections of them, times the normal vector. That is perpendicular to both of the normal unit Vector and you figure out whether it is either - which of the two perpendicular Vector it is by using the right hand rule. But what if we don’t have a visually defined, we don’t have the theta’s, the angles between them? What if for example - I will tell you that the Vector (a), what if I were to give it to you in engineering notation? An engineering notation, you essentially just breaking down the Vector into it’s x, y and z components. So let’s say the unit Vector (a) is 5i, (i) is just the unit Vector in the x direction minus 6j + 3k. These are all, I, J and K are just the unit vectors of the x, y and z directions. And the five is how much it goes in the x direction, the minus six is how much goes to the y direction and three is how much it goes in the z direction. You could try to graph it and I am trying to look for a graphing calculator that will do this so I can show you all in videos to give you more intuition. But let’s say this are your given, and let’s say that (b) - I am just making these numbers up. Let’s say its - 2i and of course we are working in three dimensions right now, plus 7j = 4k. And you could graph them but obviously if you were given a problem and actually you are trying to model Vectors in a computer simulation. This is the way you would do it, you’d break it up to the x, y and z component. Because of the add Vectors, you just have to add the respective components. But how do you multiply them? Either taking the cross or the dot product, well it actually that turns out, I am not going to prove it here, but I will just show you how to do it. That the dot product, is very easy when you have it given in this notation; and actually another way writing this notation; sometimes in bracket notations. Sometimes they would re-write this as 5 - 6, 3 where just the magnitudes of the x, y and z direction. Just want to make sure you are comfortable with all of these various notations. You could write these as -2, 7, 4, these are all the same thing. So you couldn’t daunt if you see one or the other, but anyway so how do I take (a.b)? This, I think you will find fairly pleasant. All you do is you multiply the i components add that to he j components multiplied and then add that to the k components multiplied together. So it would be 5(- 2 + - 6(7) + 3 (4), so that equals -10, -42 + 12, so this is -52 + 12 = -40 that is it. It is just a number and I should be curious to graph this on a three dimensional some type of graph to see why it is -40. There must be kind of going in opposite directions and their projections on to each other going to opposite directions. So that is why we got a minus number - the purpose of this video, I don’t want to get too much into the intuition. This is just how to calculate, where it is fairly straight forward, you just multiply the x components. Add that to the y components multiplied and add that to the z components multiplied. So whenever I am given something in engineering or bracket notation and I have to find the dot product? It’s very almost soothing and not so error prone. But as you will see, taking the cross products of these two Vectors when given in this notation isn't so straight forward. And I want you to keep in mind, you know another way you could’ve done it, you could’ve figure it out the magnitude of each of these Vectors. And then you could’ve used some fancy Trigonometry to figure out the Thetas and then used this definition. But I think you appreciate the fact that this is much simpler way of doing it. So the dot product is a lot of fun, now let’s see if we can take the cross-product. And I am not going to once again; I am not going to prove it. I am just going to show you how to do it, in the future video. I’m sure I will get a request to do it eventually and I will prove it. But the cross-product, this is more involved and I never look forward taking the cross-product of two Vectors in engineering notation. A cross B, it equals, so this is an application of matrixes. So what you do is you take the determinant on the top line of the determinant. And this is really just the way to make you memorize how you to do it, it doesn’t give you much intuition. But the intuition is given by the actual definition. If you know how much of the Vectors are perpendicular to each other, multiply those magnitudes right hand rule - it figures out what direction you are pointing in. But the way to do it if you are given in the engineering notation, you write the; i, j, k unit Vectors in the top row, then you write the first Vector in the cross-product. because order matters, so its 5 - 6, 3 that you take the second Vector which is b which -2, 7, 4. So you take the determinant of this 3x3 matrix and how do I do that? Well this is equal to the sub-determinant for i, so the sub-determinant for i, if you get rid of this column and this row. The determinant is left over, so that is -6,3,7,4 times i, I want to review the determinant if you earn by to do this but maybe me worth working through it. It will just jog your memory and then remember its plus, minus, plus so then minus the sub-determinant for j, what is the sub-determinant for j? Well you crossed out j’s row and columns, you have 5, 3 - 2, 4. We just crossed out j’s row and column and what we are left over, those are the numbers in its sub-determinant that is what I call it. J +, I always I want to do them all in one line because it will look a little bit neater. Plus the sub-determinant for k, crossed out the row and the column for k. That is 5 - 6 - 2 and seven, times k and now let’s us calculate them. And let me make some space because I have written this too vague. So what do we get? Let’s take this up here, so these 2x2 determinants are pretty easy, this is -6(4)-7(3). -24, - 21(i) - 5(4) = 20 - 2(3) so -6j + 5(7) = 35 - 12k, we could simply this which equals -24 and -21 it is -35i. And then what is 20 - -6? Well that is 20 + +26 or 26 and we have a minus out here, so -26j and those 35 - 12 = 23 + 23k so that is the cross-product. And if you were to graph this in three dimensions, you will see and this is what is interesting. You will see that this Vector, if my math is correct -35i - 26j +23k is perpendicular to both of these Vectors. Anyway I think I leave you there for now and I will see you in the next video and hopefully I can track down a Vector graphing program. Because I think it will be fun to both calculate the dot and the cross-products using these methods that I just showed you. And then to graph them and to show that it really does work that this Vector really is the right hand. It is really perpendicular to both of these and pointing in the direction as you would predict using the right hand rule. Anyway I will see you in the next video.