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‫So all that was for the vector's in three dimensions, which is exactly what we need for discourse.

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‫However, you can also compute the cross product mathematically.

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‫Let's see how to do it.

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‫Suppose that you have a vector V1 and vector V two, then you would compute the one vector, cross the

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‫two vector, or you can write it in this form.

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‫Then you would compute the cross product like this.

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‫You create a three by three matrix where the first row will be this one and these are the unit vectors.

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‫Then the second row will be the one vector.

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‫But if here it's as a column vector here you write it as a row vector and the third row will be the

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‫V two vector.

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‫But now as our role vector and by the way, the unit vectors I j and K, their purpose is to give you

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‫a direction.

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‫They're magnitudes are all one.

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‫So I has a magnitude of one, J has a magnitude of one and K has a magnitude of one.

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‫But what they do they tell you in which direction you're going.

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‫So for example, this V one vector, I can also write it down like this.

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‫So this vector here that I have written in the form of column vector, I can write it like this v one

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‫vector equals V one X I plus V one Y J and plus V one Z, K.

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‫This formulation here is equivalent to this formulation here.

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‫And so this I J and K, their unit vectors, they are all equal to one.

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‫But you know that if you see I then you know that this component is in the X direction and thanks to

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‫J you know that this component is in the Y direction.

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‫And thanks to K you know that this component is in the Z direction.

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‫And we're going to talk about the unit vectors more later in this section as well.

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‫And so let's go back to computing our cross product.

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‫We have this three by three matrix that we have constructed like this.

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‫And now what we do, we take the determinant of it, which looks like this here.

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‫I list the I J and K unit vectors horizontally and I assign the X components of the V one and the two

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‫vectors to I like this then the Y components of these two vectors I assigned to J and the Z components

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‫of them I assign to K and then I compute the determinant of the matrix.

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‫The formula for the determinant comes from linear algebra and it looks like this.

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‫You take the eye vector and you cross the corresponding row and column and the two by two matrix that

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‫is remaining, you take its determinant and that will be your eye or X component in your total cross

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‫product.

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‫So it will be like that.

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‫You take the eye vector and then to compute the determinant of a two by two matrix, you first multiply

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‫the one Y times V to Z and then you subtract V, one Z times V to Y.

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‫Let's just call it a four now and then you do the same thing for J and you cross out the corresponding

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‫row and column your remaining two by two matrix will be this one composed of these elements here.

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‫But in case of J there is an exception.

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‫You have to put minus in front of J.

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‫So the J component of your total cross product will be this one, this entire thing.

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‫And let's call this part B, and now the only thing that's left is to find the key part of this total

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‫cross product, and that will be your exercise.

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‫Why don't you find the key part for this cross product and you'll see the solution in the next video?

