If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Cross product 2

A little more intuition on the cross product. Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user Hafizul Bari
    in cross product why we use Sine, but not Cosine
    (29 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user TheSupremeOverLord
      The primary purpose of "cross product" is to calculate areas. If you have two vectors, u⃗ and v⃗ , the area of the parallelogram having those two vectors as two sides is, of course, "base times height". The "base" is the length of one of the vector, |u⃗ |, say. To Find the "height", draw a line from the tip of v⃗ perpendicular to u⃗ . That gives a right triangle having |v⃗ | as hypotenuse. The "height" is the length of the "opposite side" which is given by |v⃗ |sin(θ) where θ is the angle between the vectors. That is, the area of the parallelogram formed by vectors u⃗ and v⃗ is |u⃗ ||v⃗ |sin(θ).

      A primary use of the dot product, on the other hand, is to find the projection of one vector on the other. Again, draw a line from the tip of u⃗ perpedicular to v⃗ . The projection of u⃗ onto v⃗ is now the near side of the right triangle produced. Its length is |u|cos(θ). To find the actual vector projection, multiply that length by a unit vector in the direction of v⃗ which is v⃗ /|v⃗ . That is, the vector projection of u⃗ on v⃗ is given by |u⃗ |cos(θ)[v⃗ /|v⃗ |] which we can simplify by multiplying both numerator and denominator by |v⃗ |:
      |u⃗ ||u⃗ |cos(θ)|v⃗ |2v⃗

      The numerator of that fraction is u⃗ ⋅v⃗ .

      I have done the vectors and other math symbols by using "LaTeX" which is implemented on this board To see the code used, click on the formula or click on "quote". There are several tutorials on the internet.
      (11 votes)
  • blobby green style avatar for user arman91
    This isn't intuitive at all for me. If at you do exactly what he said (index finger points in the direction of b and middle finger for a) with your right hand, i still get my thumb going into the screen, the same result as in the last video. How is he flexing his thumb to get it to point out of the page? I'm really confused.
    (19 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Daniel Kurniawan
    if we multiply the vectors as a cross product, why do we get another different vector, not the area?

    I really don't get the intuition... :S
    (6 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user michel
      This is how I answered another similar question, hope it helps you as well...
      We're not really finding an area when we are multiplying 2 vectors. You may get that idea because when we usually multiply two lengths in basic maths, we get the area that those two lengths form between them. However, those two lengths are scalars, and not vectors (which have directions). So when you multiply vectors, not only are you multiplying the magnitudes, you are also multiplying the directions. It may not be intuitive, but that's how multiplying vectors work. The "getting an area from two values" idea only applies to multiplying two scalars. Hope this helps.
      (3 votes)
  • blobby green style avatar for user abdullaaldilaijan
    I asked this to my teacher in school, but i didn't get a convincing answer. If the multiplying 2 vectors with cross product leads you to the third dimension, what happens when you multiply 3 vectors with cross product? a fellow student said out of the bloom that it would lead back to the plane, which i found very convincing, but my teacher didn't agree and kept saying other stuff uncomprehensively (that's why i love khan academy).. what is the right answer?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user abdullaaldilaijan
      so you agree that it's always gonna rotate between the 3 axes(planes)?

      but in your example i noticed that the degree between angles was 90.. an interesting idea i thought about: if for example we have a and b as vectors. and the degree between them is 45. a being on the x axis, and b between x and y axes. "a x b" would result with c wich would be parallel/on to the z axis. Now if you multiply "c x a" you'd return to the y axis which is perpendicular to both because one is parallel to x-axis and the other to z-axis. however if you multiply "c x b" the plane that both lie on in the same time is the a new plane with axis with (mathematically speaking) x=y is the axis and the result would probably give us a new axis which is (if you put the ordinary plane of x, y in front of you) of degree 135,,, and you could go on circuling in each direction, and you'd probably get 8 vectors on each plane... :\

      just a thought
      (2 votes)
  • blobby green style avatar for user DillonWu999
    why do we use the right hand rule? i mean, im sure a guy didn't just say oh, the direction of the third vector can be found with our right hand, without having an reasoning to support it
    (3 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Miles Ford
      The point is that the three fingers of your right hand form the three axes of a 3D Cartesian system. Any two vectors lie on a single plane. The cross product finds a vector that is perpendicular to the plane, or perpendicular to both the vectors you're finding the cross product of. Your fingers just give you a visual representation. There are actually two vectors that are perpendicular, the one at 180 degrees to the one you find, which you'd get if you used the left hand. So, why the right hand? Just convention, it's got to be one of them... Why drive on the left or right side of the road? Neither is correct, it's just important everyone does the same.
      (6 votes)
  • hopper cool style avatar for user Iron Programming
    Can I get an example of why knowing the direction (in vs out) of the cross-product vector is important. Also, I am sorry to say that I don't think that I understand Sal's explanation of the right hand rule at all. I'll need to see a video of someone using their actually hand before I can understand it.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user loveapplejuice888
    is it always true that pointing up (out of page), and pointing down (into the page)? i am confused as to how Sal can tell it is into or out of page, when is is pointing up or down.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user haadimahmood
    At |a|opp/|A| is confusing .|A| is a vector not a scalar.So how can you just convert a vector into scalar?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • marcimus pink style avatar for user cara.ann.pomeroy
      When we take the absolute value of a vector it give us just the magnitude of the vector. That's intuitive; think what we mean when we're taking absolute values of pure numbers: We're figuring out the distance that our number is from our origin; it doesn't matter the direction (negative or positive). In a vector, we have infinite possibilities for direction. When we take the absolute value, we ignore the direction and take only the distance from the origin. When we're talking about vectors, that distance is what we mean when we say "magnitude". And magnitude is a scalar. So lAl is just another way of saying "the magnitude of A".
      (0 votes)
  • blobby green style avatar for user naomic.crowder
    How can you use this if the vectors are given in ijk form and there is no angle given
    (0 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Ali
      I think if you do this graphically using a matrix, it's much easier to remember. I'll try to create one here. So you need to find the determinant of a 3x3 matrix. Let's say vector A=<Ax,Ay,Az> and vector B=<Bx,By,Bz>. Unit direction i, j, and k correspond to x, y, and z.

      In this matrix, the first row is supposed to have a ^ on top of i,j, and k. I'm not sure if the formatting will come through.

      | i^ j^ k^ |
      | Ax Ay Az |
      | Bx By Bz |

      To find the determinant, you need to use sub-determinants. For example, starting with i^, you cross out the row and the column that contains i^ and multiply it by the 2x2 determinant left in this case:
      |Ay Az|
      |By Bz|
      Same idea for j^, but there IS A NEGATIVE SIGN in front of j^ and you cross out the row and column that contain j^. You can find k^ the same way as i^, but crossing out the row and column that contain k^. The determinant of the 2x2 matrix gives you the magnitude and the unit vectors give the components of the vector.
      (1 vote)
  • blobby green style avatar for user jonpdw
    How do you decide what vector to put first when using the cross product in equations?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

Let's see if we can get a little bit more practice and intuition of what cross products are all about. So in the last example, we took a cross b. Let's see what happens when we take b cross a. So let me erase some of this. I don't want to erase all of it because it might be useful to give us some intuition to compare. I'm going to keep that. Actually, I can erase this, I think. So the things I have drawn here, this was a cross b. Let me cordon it off so you don't get confused. So that was me using the right hand rule when I tried to do a cross b, and then we saw that the magnitude of this was 25, and n, the direction, pointed downwards. Or when I drew it here, it would point into the page. So let's see what happens with b cross a, so I'm just switching the order. b cross a. Well, the magnitude is going to be the same thing, right? Because I'm still going to take the magnitude of b times the magnitude of a times the sine of the angle between them, which was pi over 6 radians and then times some unit vector n. But this is going to be the same. When I multiply scalar quantities, it doesn't matter what order I multiply them in, right? So this is still going to be 25, whatever my units might have been, times some vector n. And we still know that that vector n has to be perpendicular to both a and b, and now we have to figure out, well, is it, in being perpendicular, it can either kind of point into the page here or it could pop out of the page, or point out of the page. So which one is it? And then we take our right hand out, and we try it again. So what we do is we take our right hand. I'm actually using my right hand right now, although you can't see it, just to make sure I draw the right thing. So in this example, if I take my right hand, I take the index finger in the direction of b. I take my middle finger in the direction of a, so my middle figure is going to look something like that, right? And then I have two leftover fingers there. Then the thumb goes in the direction of the cross product, right? Because your thumb has a right angle right there. That's the right angle of your thumb. So in this example, that's the direction of a, this is the direction of b, and we're doing b cross a. That's why b gets your index finger. The index finger gets the first term, your middle finger gets the second term, and the thumb gets the direction of the cross product. So in this example, the direction of the cross product is upwards. Or when we're drawing it in two dimensions right here, the cross product would actually pop out of the page for b cross a. So I'll draw it over. It would be the circle with the dot. Or if I were to draw it analogous to this, so this right here, that was a cross b. And then b cross a is the exact same magnitude, but it goes in the other direction. That's b cross a. It just flips in the opposite direction. And that's why you have to use your right hand, because you might know that, oh, something's going to pop in or out of the page, et cetera, et cetera, but you need to know your right hand to know whether it goes in or out of the page. Anyway, let's see if we can get a little bit more intuition of what this is all about because this is all about intuition. And frankly, I'll tell you, the cross product comes into use in a lot of concepts that frankly we don't have a lot of real-life intuition, with electrons flying through a magnetic field or magnetic fields through a coil. A lot of things in our everyday life experience, maybe if we were metal filings living in a magnetic field-- well, we do live in a magnetic field. In a strong magnetic field, maybe we would get an intuition, but it's hard to have as deep of an intuition as we do for, say, falling objects, or friction, or forces, or fluid dynamics even, because we've all played with water. But anyway, let's get a little bit more intuition. And let's think about why is there that sine of theta? Why not just multiply the magnitudes times each other and use the right hand rule and figure out a direction? What is that sine of theta all about? I think I need to clear this up a little bit just so this could be useful. So why is that sine of theta there? Let me redraw some vectors. I'll draw them a little fatter. So let's say that's a, that's a, this is b. b doesn't always have to be longer than a. So this is a and this is b. Now, we can think of it a little bit. We could say, well, this is the same thing as a sine theta times b, or we could say this is b sine theta times a. I hope I'm not confusing-- all I'm saying is you could interpret this as-- because these are just magnitudes, right? So it doesn't matter what order you multiply them in. You could say this is a sine theta times the magnitude of b, all of that in the direction of the normal vector, or you could put the sine theta the other way. But let's think about what this would mean. a sine theta, if this is theta. What is a sine theta? Sine is opposite over hypotenuse, right? So opposite over hypotenuse. So this would be the magnitude of a. Let me draw something. Let me draw a line here and make it a real line. Let me draw a line there, so I have a right angle. So what's a sine theta? This is the opposite side. So a sine theta is a, and sine of theta is opposite over hypotenuse. The hypotenuse is the magnitude of a, right? So sine of theta is equal to this side, which I call o for opposite, over the magnitude of a. So it's opposite over the magnitude of a. So this term a sine theta is actually just the magnitude of this line right here. Another way you could-- let me redraw it. It doesn't matter where the vectors start from. All you care about is this magnitude and direction, so you could shift vectors around. So this vector right here, and you could call it this opposite vector, that's the same thing as this vector. That's the same thing as this. I just shifted it away. And so another way to think about it is, it is the component of vector a, right? We're used to taking a vector and splitting it up into x- and y-components, but now we're taking a vector a, and we're splitting it up into-- you can think of it as a component that's parallel to vector b and a component that is perpendicular to vector b. So a sine theta is the magnitude of the component of vector a that is perpendicular to b. So when you're taking the cross product of two numbers, you're saying, well, I don't care about the entire magnitude of vector a in this example, I care about the magnitude of vector a that is perpendicular to vector b, and those are the two numbers that I want to multiply and then give it that direction as specified by the right hand rule. And I'll show you some applications. This is especially important-- well, we'll use it in torque and we'll also use it in magnetic fields, but it's important in both of those applications to figure out the components of the vector that are perpendicular to either a force or a radius in question. So that's why this cross product has the sine theta because we're taking-- so in this, if you view it as magnitude of a sine theta times b, this is kind of saying this is the magnitude of the component of a perpendicular to b, or you could interpret it the other way. You could interpret it as a times b sine theta, right? Put a parentheses here. And then you could view it the other way. You could say, well, b sine theta is the component of b that is perpendicular to a. Let me draw that, just to hit the point home. So that's my a, that's my b. This is a, this is b. So b has some component of it that is perpendicular to a, and that is going to look something like-- well, I've run out of space. Let me draw it here. If that's a, that's b, the component of b that is perpendicular to a is going to look like this. It's going to be perpendicular to a, and it's going to go that far, right? And then you could go back to SOH CAH TOA and you could prove to yourself that the magnitude of this vector is b sine theta. So that is where the sine theta comes from. It makes sure that we're not just multiplying the vectors. It makes sure we're multiplying the components of the vectors that are perpendicular to each other to get a third vector that is perpendicular to both of them. And then the people who invented the cross product said, well, it's still ambiguous because it doesn't tell us-- there's always two vectors that are perpendicular to these two. One goes in, one goes out. They're in opposite directions. And that's where the right hand rule comes in. They'll say, OK, well, we're just going to say a convention that you use your right hand, point it like a gun, make all your fingers perpendicular, and then you know what direction that vector points in. Anyway, hopefully, you're not confused. Now I want you to watch the next video. This is actually going to be some physics on electricity, magnetism and torque, and that's essentially the applications of the cross product, and it'll give you a little bit more intuition of how to use it. See you soon.