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  • Hey folks! Where we left off,

  • I was talking about how to compute a three-dimensional cross product

  • between two vectors, v x w.

  • It's this funny thing where you write a matrix, whose second column has the coordinates of

  • v,

  • whose third column has the coordinates of w,

  • but the entries of that first column, weirdly, are the symbols i-hat, j-hat and k-hat

  • where you just pretend like those guys are numbers for the sake of computations.

  • Then with that funky matrix in hand,

  • you compute its determinant.

  • If you just chug along with those computations, ignoring the weirdness,

  • you get some constant times i-hat + some constant times j-hat + some constant times k-hat.

  • How specifically you think about computing that determinant

  • is kind of beside the point.

  • All that really matters here is that you'll end up with three different numbers

  • that are interpreted as the coordinates of some resulting vector.

  • From here, students are typically told to just believe that

  • the resulting vector has the following geometric properties.

  • Its length equals the area of the parallelogram defined by v and w.

  • It points in a direction perpendicular to both of v and w.

  • And this direction obeys the right hand rule

  • in the sense that if you point your forefinger along v

  • and your middle finger along w

  • then when you stick up your thumb

  • it'll point in the direction of the new vector.

  • There are some brute force computations

  • that you could do to confirm these facts.

  • But I want to share with you a really elegant line of reasoning.

  • It leverages a bit of background, though.

  • So for this video I'm assuming that everybody has watched chapter 5 on the determinant

  • and chapter 7 where I introduce the idea of duality.

  • As a quick reminder, the idea of duality is that

  • anytime you have a linear transformation from some space to the number line,

  • it's associated with a unique vector in that space

  • in the sense that performing the linear transformation

  • is the same as taking a dot product with that vector.

  • Numerically, this is because one of those transformations

  • is described by a matrix with just one row

  • where each column tells you the number that each basis vector lands on.

  • And multiplying this matrix by some vector v is computationally identical to

  • taking the dot product between v and the vector you get by turning that matrix on its side.

  • The takeaway is that whenever you're out in the mathematical wild

  • and you find a linear transformation to the number line

  • you will be able to match it to some vector

  • which is called thedual vectorof that transformation

  • so that performing the linear transformation

  • is the same as taking a dot product with that vector.

  • The cross product gives us a really slick example of this process in action.

  • It takes some effort, but it's definitely worth it.

  • What I'm going to do is to define a certain linear transformation from three dimensions

  • to the number line.

  • And it will be defined in terms of the two vectors v and w.

  • Then, when we associate that transformation with itsdual vectorin 3D space

  • thatdual vectoris going to be the cross product of v and w.

  • The reason for doing this will be that understanding that transformation

  • is going to make clear the connection between the computation and the geometry of the cross

  • product.

  • So to back up a bit,

  • remember in two dimensions what it meant to compute the 2D version of the cross product?

  • When you have two vectors v and w,

  • you put the coordinates of v as the first column of the matrix

  • and the coordinates of w is the second column of matrix

  • then you just compute the determinant.

  • There's no nonsense with basis vectors stuck in a matrix or anything like that.

  • Just an ordinary determinant returning a number.

  • Geometrically, this gives us the area of a parallelogram

  • spanned out by those two vectors

  • with the possibility of being negative, depending on the orientation of the vectors.

  • Now, if you didn't already know the 3D cross product

  • and you're trying to extrapolate

  • you might imagine that it involves taking three separate 3D vectors u, v and w.

  • And making their coordinates the columns of a 3x3 matrix

  • then computing the determinant of that matrix.

  • And, as you know from chapter 5

  • geometrically, this would give you the volume of a parallelepiped

  • spanned out by those three vectors

  • with the plus or minus sign

  • depending on the right-hand rule orientation of those three vectors.

  • Of course, you all know that this is not the 3D cross product.

  • The actual 3D cross product takes in two vectors and spits out a vector.

  • It doesn't take in three vectors and spit out a number.

  • But this idea actually gets us really close to what the real cross product is.

  • Consider that first vector u to be a variable

  • say, with variable entries x, y and z

  • while v and w remain fixed.

  • What we have then is a function from three dimensions to the number line.

  • You input some vector x, y, z and you get out a number

  • by taking the determinant of a matrix whose first column is x, y, z

  • and whose other two columns are the coordinates of the constant vectors v and w.

  • Geometrically, the meaning of this function is that

  • for any input vector x, y, z, you consider the parallelepiped defined by this vector

  • v and w

  • then you return its volume with the plus or minus sign depending on orientations.

  • Now, this might feel like kind of a random thing to do.

  • I mean, where does this function come from?

  • Why are we defining it this way?

  • And I'll admit at this stage of my kind of feel like it's coming out of the blue.

  • But if you're willing to go along with it

  • and play around with the properties that this guy has

  • it's the key to understanding the cross product.

  • One really important fact about this function is that it's linear.

  • I'll actually leave it to you to work through the details of why this is true

  • based on properties of the determinant.

  • But once you know that it's linear

  • we can start bringing in the idea ofduality”.

  • Once you know that it's linear

  • you know that there's some way to describe this function as matrix multiplication.

  • Specifically, since it's a function that goes from three dimensions to one dimension

  • there will be a 1x3 matrix that encodes this transformation.

  • And the whole idea of duality

  • is that the special thing about transformations from several dimensions to one dimension

  • is that you can turn that matrix on its side

  • and, instead, interpret the entire transformation as the dot product with a certain vector.

  • What we're looking for is the special 3D vector that I'll call p

  • such that taking the dot product between p and any other vector [x, y, z]

  • gives the same result as plugging in [x, y, z] as the first column of a 3x3 matrix

  • whose other two columns have the coordinates of v and w

  • then computing the determinant.

  • I'll get to the geometry of this in just a moment.

  • But right now, let's dig in and think about what this means computationally.

  • Taking the dot product between p and [x, y, z]

  • will give us something times x + something times y + something times z

  • where those somethings are the coordinates of p.

  • But on the right side here, when you compute the determinant

  • you can organize it to look like some constant times x + some constant times y + some constant

  • times z

  • where those constants involve certain combinations of the components of v and w.

  • So, those constants, those particular combinations of the coordinates of v and w

  • are going to be the coordinates of the vector p that we're looking for.

  • But what's going on the right here

  • should feel very familiar to anyone

  • who's actually worked through a cross-product computation.

  • Collecting the constant terms that are multiplied by x, y and z like this

  • is no different from plugging in the symbols i-hat, j-hat and k-hat to that first column

  • and seeing which coefficients aggregate on each one of those terms.

  • It's just that plugging in i-hat, j-hat and k-hat

  • is a way of signaling that we should interpret those coefficients as the coordinates of a

  • vector.

  • So, what all of this is saying

  • is that this funky computation can be thought of as a way to answer the following question:

  • What vector p has the special property

  • that when you take a dot product between p and some vector [x, y, z]

  • it gives the same result as plugging in [x, y, z] to the first column of the matrix

  • whose other two columns have the coordinates of v and w

  • then computing the determinant?

  • That's a bit of a mouthful.

  • But it's an important question to digest for this video.

  • Now for the cool part which ties all this together

  • with the geometric understanding of the cross product that I introduced last video.

  • I'm going to ask the same question again.

  • But this time, we're going to try to answer it geometrically

  • instead of computationally.

  • What 3D vector p has the special property

  • that when you take a dot product between p and some other vector [x, y, z]

  • it gives the same result as if you took the signed volume of a parallelepiped

  • defined by this vector [x, y, z] along with v and w?

  • Remember, the geometric interpretation of a dot product

  • between a vector p and some other vector

  • is to project that other vector onto p

  • then to multiply the length of that projection by the length of p.

  • With that in mind, let me show a certain way to think about

  • the volume of the parallelepiped that we care about.

  • Start by taking the area of the parallelogram defined by v and w

  • then multiply it, not by the length of [x, y, z]

  • but by the component of [x, y, z] that's perpendicular to that parallelogram.

  • In other words, the way our linear function works on a given vector

  • is to project that vector onto a line that's perpendicular to both v and w

  • then, to multiply the length of that projection by the area of the parallelogram spanned by