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Last video, I've talked about the dot product.
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Showing both the standard introduction to the topic,
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as well as a deeper view of how it relates to linear transformations.
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I'd like to do the same thing for cross products,
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which also have a standard introduction
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along with a deeper understanding in the light of linear transformations.
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But this time I am dividing it into two separate videos.
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Here i'll try to hit the main points
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that students are usually shown about the cross product.
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And in the next video,
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I'll be showing a view which is less commonly taught, but really satisfying when you learn
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it.
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We'll start in two dimensions.
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If you have two vectors v̅ and w̅,
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think about the parallelogram that they span out
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What i mean by that is,
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that if you take a copy of v̅
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and move its tail to the tip of w̅,
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and you take a copy of w̅
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And move its tail to the tip of v̅,
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the four vectors now on the screen enclose a certain parallelogram.
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The cross product of v̅ and w̅,
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written with the X-shaped multiplication symbol,
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is the area of this parallelogram.
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Well, almost. We also need to consider
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orientation. Basically, if v̅ is on the
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right of w̅, then v̅×w̅ is positive
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and equal to the area of the
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parallelogram. But if v̅ is on the left of w̅,
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then the cross product is negative,
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namely the negative area of that
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parallelogram. Notice this means that
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order matters. If you swapped v̅ and w̅
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instead taking w̅×v̅, the cross
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product would become the negative of
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whatever it was before. The way I always
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remember the ordering here is that when
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you take the cross product of the two
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basis vectors in order, î×ĵ,
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the results should be positive. In fact,
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the order of your basis vectors is what
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defines orientation so since î is on
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the right of ĵ, I remember that v̅×w̅
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has to be positive whenever v̅ is
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on the right of w̅.
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So, for example with the vector shown
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here, I'll just tell you that the area of
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that parallelogram is 7. And since v̅
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is on the left of w̅, the cross product
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should be negative so v̅×w̅ is -7.
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But of course you want to be able to
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compute this without someone telling you
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the area. This is where the determinant comes in.
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So, if you didn't see Chapter 5 of this
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series, where I talk about the
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determinant now would be a really good
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time to go take a look.
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Even if you did see it, but it was a while
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ago. I'd recommend taking another look
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just to make sure those ideas are fresh in your mind.
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For the 2-D cross-product v̅×w̅,
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what you do is you write the coordinates
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of v̅ as the first column of the matrix
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and you take the coordinates of w̅ and
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make them the second column then you
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just compute the determinant.
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This is because a matrix whose columns
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represent v̅ and w̅ corresponds with a
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linear transformation that moves the
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basis vectors î and ĵ to v̅ and w̅.
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The determinant is all about measuring
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how areas change due to a transformation.
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And the prototypical area that we look
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at is the unit square resting on î and ĵ.
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After the transformation,
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that square gets turned into the
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parallelogram that we care about.
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So the determinant which generally
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measures the factor by which areas are
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changed, gives the area of this
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parallelogram; since it evolved from a
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square that started with area 1.
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What's more if v̅ is on the left of w̅, it
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means that orientation was flipped
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during that transformation, which is what
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it means for the determinant to be negative.
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As an example let's say v̅ has
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coordinates negative (-3,1) and w̅ has
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coordinates (2,1). The determinant of the
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matrix with those coordinates as columns
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is (-3·1) - (2·1),
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which is -5. So evidently the
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area of the parallelogram we define is 5
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and since v̅ is on the left of w̅, it
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should make sense that this value is
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negative. As with any new operation you learn
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I'd recommend playing around with this
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notion of it in your head just to get
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kind of an intuitive feel for what the
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cross product is all about.
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For example you might notice that when
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two vectors are perpendicular or at
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least close to being perpendicular their
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cross product is larger than it would be
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if they were pointing in very similar
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directions. Because the area of that
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parallelogram is larger when the sides
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are closer to being perpendicular.
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Something else you might notice is that
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if you were to scale up one of those
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vectors, perhaps multiplying v̅ by three
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then the area of that parallelogram is
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also scaled up by a factor of three.
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So what this means for the operation is
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that 3v̅×w̅ will be exactly three
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times the value of v̅×w̅ .
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Now, even though all of this is a
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perfectly fine mathematical operation
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what i just described is technically not
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the cross-product. The true cross product
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is something that combines two different
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3D vectors to get a new 3D vector. Just as before,
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we're still going to consider the
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parallelogram defined by the two vectors
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that were crossing together. And the area
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of this parallelogram is still going to
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play a big role. To be concrete let's say
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that the area is 2.5 for the vectors
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shown here but as I said the cross
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product is not a number it's a vector.
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This new vector's length will be the area
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of that parallelogram which in this case
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is 2.5. And the direction of that new
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vector is going to be perpendicular to
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the parallelogram. But which way!, right?
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I mean there are two possible vectors with
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length 2.5 that are perpendicular to a given plane.
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This is where the right hand rule comes
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in. Put the fore finger of your right hand
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in the direction of v̅ then stick out
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your middle finger in the direction of w̅.
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Then when you point up your thumb, that's the
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direction of the cross product.
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For example let's say that v̅ was a
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vector with length 2 pointing straight
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up in the Z direction and w̅ is a vector
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with length 2 pointing in the pure Y
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direction. The parallelogram that they
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define in this simple example is
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actually a square, since they're
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perpendicular and have the same length.
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And the area of that square is 4. So
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their cross product should be a vector
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with length 4. Using the right hand
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rule, their cross product should point in the negative X direction.
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So the cross product of these two
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vectors is -4·î.
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For more general computations,
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there is a formula that you could
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memorize if you wanted but it's common
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and easier to instead remember a certain
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process involving the 3D determinant.
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Now, this process looks truly strange at
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first. You write down a 3D matrix where
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the second and third columns contain the
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coordinates of v̅ and w̅. But for that
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first column you write the basis vectors
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î, ĵ and k̂. Then you compute
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the determinant of this matrix. The
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silliness is probably clear here.
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What on earth does it mean to put in a
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vector as the entry of a matrix?
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Students are often told that this is
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just a notational trick. When you carry
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out the computations as if î, ĵ and k̂
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were numbers, then you get some
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linear combination of those basis vectors.
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And the vector
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defined by that linear combination, students
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are told to just believe, is the unique
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vector perpendicular to v̅ and w̅ whose
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magnitude is the area of the appropriate
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parallelogram and whose direction obeys
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the right hand rule.
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And, sure!. In some sense this is just a
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notational trick. But there is a reason
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for doing in.
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It's not just a coincidence that the
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determinant is once again important. And
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putting the basis vectors in those slots
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is not just a random thing to do. To
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understand where all of this comes from
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it helps to use the idea of duality that
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I introduced in the last video.
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This concept is a little bit heavy
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though, so I'm putting it in a separate
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follow-on video for any of you who are
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curious to learn more.
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Arguably it falls outside the essence of
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linear algebra. The important part here
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is to know what that cross product
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vector geometrically represents. So if
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you want to skip that next video, feel
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free. But for those of you who are
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willing to go a bit deeper and who are
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curious about the connection between
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this computation and the underlying
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geometry, the ideas that I will talk about
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in the next video or just a really
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elegant piece of math.