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# Cross product 2

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

## Want to join the conversation?

• in cross product why we use Sine, but not Cosine • 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.
• 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. • 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 • 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.
• 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? • 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
• 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 • 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.
• 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. • 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. • At |a|opp/|A| is confusing .|A| is a vector not a scalar.So how can you just convert a vector into scalar? • 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".
• How can you use this if the vectors are given in ijk form and there is no angle given • 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)
• How do you decide what vector to put first when using the cross product in equations?
(1 vote) 