From b7aeb6b1f8e6fd620b476bf348fb1c5754ec1223 Mon Sep 17 00:00:00 2001 From: Kurt Bruns <12160501+kurtbruns@users.noreply.github.com> Date: Sat, 13 Jan 2024 13:51:36 -0800 Subject: [PATCH] Publish cross-products-extended (#387) Co-authored-by: Kurt Bruns --- .../QuestionCrossProduct1Answer.txt | 47 ++ .../conclusion/3dTo1dLinearTransformation.png | 3 + .../conclusion/3dTo1dLinearTransformation.svg | 229 ++++++++ .../conclusion/CrossProductFormula.svg | 1 + .../DualVectorComputationalApproach.png | 3 + .../DualVectorComputationalApproach.svg | 226 ++++++++ .../DualVectorGeometricApproach.png | 3 + .../DualVectorGeometricApproach.svg | 475 ++++++++++++++++ .../conclusion/DualVectorOfTransformation.png | 3 + .../conclusion/DualVectorOfTransformation.svg | 272 +++++++++ .../CrossProductParallelogram.png | 3 + .../CrossProductParallelogram.svg | 1 + .../NumericalFormulaAndFactoids.png | 3 + .../figures/introduction/RightHandRule.png | 3 + .../figures/introduction/RightHandRule.svg | 275 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public/content/lessons/2016/cross-products/figures/{ => two-dimensions}/two-dimensions-area-of-parallelogram.png (100%) rename public/content/lessons/2016/cross-products/figures/{ => two-dimensions}/two-dimensions-cross-product-negative.png (100%) rename public/content/lessons/2016/cross-products/figures/{ => two-dimensions}/two-dimensions-cross-product-positive.png (100%) diff --git a/public/content/lessons/2016/cross-products-extended/QuestionCrossProduct1Answer.txt b/public/content/lessons/2016/cross-products-extended/QuestionCrossProduct1Answer.txt new file mode 100644 index 00000000..faf9283e --- /dev/null +++ b/public/content/lessons/2016/cross-products-extended/QuestionCrossProduct1Answer.txt @@ -0,0 +1,47 @@ +$$ +\left[\begin{array}{l} +v_1 \\ +v_2 \\ +v_3 +\end{array}\right] \times\left[\begin{array}{l} +w_1 \\ +w_2 \\ +w_3 +\end{array}\right]=\left[\begin{array}{l} +v_2 \cdot w_3-w_2 \cdot v_3 \\ +v_3 \cdot w_1-w_3 \cdot v_1 \\ +v_1 \cdot w_2-w_1 \cdot v_2 +\end{array}\right] +$$ + +$$ +\left[\begin{array}{c} +0 \\ +2 \\ +-1 +\end{array}\right] +\times +\left[\begin{array}{c} +-2 \\ +-1 \\ +1 +\end{array}\right] += +\left[\begin{array}{l} +2-1 \\ +2-0 \\ +0+4 +\end{array}\right] += +\left[\begin{array}{l} +1 \\ +2 \\ +4 +\end{array}\right] +$$ + +This also intuitiveley makes sense because using the right-hand rule, we know that the resulting vector should end up somewhere in the first quadrant and positive $z$ direction. + +
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-1,10 +1,256 @@ --- title: Cross products in the light of linear transformations -description: The formula for the cross product can feel like a mystery, or some kind of crazy coincidence. But it isn't. There is a fundamental connection between the cross product and determinants. +description: The formula for the cross product can feel like a mystery, or some kind of crazy coincidence. But it isn't. There is a fundamental connection between the cross product and determinants. date: 2016-09-01 chapter: 11 video: BaM7OCEm3G0 source: _2016/eola/chapter8.py credits: - Lesson by Grant Sanderson +- Text adaptation by Kurt Bruns +- Text adaptation by James Schloss --- + +> "From [Grothendieck], I have also learned not to take glory in the difficulty of a proof: difficulty means we have not understood. The idea is to be able to paint a landscape in which the proof is obvious." +> +> \- Pierre Deligne + +In the last chapter, we talked about how to compute a three-dimensional cross product of two vectors, $\vec{\mathbf{v}} \times \vec{\mathbf{w}}$. It's this funny thing where you write a matrix whose second column has the coordinates of $\vec{\mathbf{v}}$, whose third columns has the coordinates of $\vec{\mathbf{w}}$, but the entries of the first column, weirdly, are the basis vectors $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$, where you pretend they are numbers for the sake of computations. + +$$ +\left[\begin{array}{l}v_1 \\v_2 \\v_3\end{array}\right] \times\left[\begin{array}{l}w_1 \\w_2 \\w_3\end{array}\right]=\operatorname{det}\left(\left[\begin{array}{lll}\hat{\imath} & v_1 & w_1 \\\hat{\jmath} & v_2 & w_2 \\\hat{k} & v_3 & w_3\end{array}\right]\right) +$$ + +If you just chug along with the computation, ignoring this weirdness, you get some constant times $\hat{\imath}$, plus some constant times $\hat{\jmath}$, plus some constant times $\hat{k}$, which defines a new 3d vector. + +$$ +\vec{\mathbf{p}} = \hat{\imath} \underbrace{\left(v_2 w_3-v_3 w_2\right)}_{\text {Some number }}+\hat{\jmath} \underbrace{\left(v_3 w_1-v_1 w_3\right)}_{\text {Some number }}+\hat{k} \underbrace{\left(v_1 w_2-v_2 w_1\right)}_{\text {Some number }} +$$ + +
+ +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 $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. +- It points in a direction perpendicular to $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, +- This direction obeys the right-hand rule, in the sense that if you point your forefinger along $\vec{\mathbf{v}}$, and your middle finger along $\vec{\mathbf{w}}$, then stick out your thumb, it will point in the direction of the new vectors. + +
+ +There are some brute-force computational ways to confirm these facts, but I want to share with you a really elegant line of reasoning. This leverages a bit of background, though, so I'm assuming everyone has read chapter 5 on the determinant, and chapter 9 where I introduce the idea of duality. So, you know, go back and take a look if needed. + +## Under the light of transformations + +As a reminder, the idea of duality is that anytime you have a linear transformation from some space to the number line, it is 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, it's because one of those transformations is described by a matrix with just one row, where each column tells you which number the basis vectors land on. And multiplying this matrix by some vector $\vec{\mathbf{v}}$ is computationally identical to taking the dot product between $\vec{\mathbf{v}}$ and the vector you get by turning that matrix on its side. + +
+ +The takeaway is that when 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 the "dual vector" of the transformation, so that performing that linear transformation is the same as taking the dot product with that vector. + +## The idea + +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 define a certain linear transformation from three dimensions to the number line, and it will be defined in terms of two vectors $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. Then, when we associate that transformation with its dual vector in 3d space, that dual vector will be the cross product of $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. Understanding that transformation will make clear the connection between the geometry and the computation of the cross product. + +
+ +To back up a bit, remember that in two dimensions, computing the 2d version of the cross product of vectors $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ involves taking the determinant of a matrix whose columns contain the coordinates of those vectors. There's no nonsense with basis vectors stuck in the matrix, just an ordinary determinant returning a number. Geometrically, this gives us the area of the parallelogram spanned out by those two vectors, with the possibility of being negative depending on the orientation of the vectors. + +
+ +If you didn't already know the 3d cross product, you might imagine that it involves taking 3 separate 3d vectors, $\vec{\mathbf{u}}$, $\vec{\mathbf{v}}$, and $\vec{\mathbf{w}}$, making their coordinates the columns of a 3x3 matrix, and computing the determinant of that matrix. As you know from chapter 5, geometrically this would give the volume of a parallelepiped spanned out by those 3 vectors, with the plus or minus sign of your result depending on their right-hand rule orientation of the three vectors. + +
+ +Of course, you all know this is not the 3d cross product, since the actual 3d cross product takes 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 the first vector $\vec{\mathbf{u}}$ to be a variable, while $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ are fixed. + +
+ +What we have, then, is a function from three dimensions to the number line. You input some vector $\vec{\mathbf{u}}$, and you get a number by taking the determinant of a matrix whose first column is $\vec{\mathbf{u}}$, and whose other two columns are the constant vectors $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. Geometrically, the meaning of this function is that for any input vector $\vec{\mathbf{u}}$, you consider the parallelepiped defined by this vector, $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, then return its volume, with a plus or minus sign depending on orientation. What's more, this function is linear. + +
+ +So here we are, out in the mathematical wild with a linear transformation that outputs numbers. The idea of duality should be perking up in your mind. What we're going to do is find the matrix that describes this transformation, which will be a 1x3 matrix since the transformation goes from three dimensions to one. In other words, we're looking for a 1x3 matrix such that multiplying this matrix by some vector $\vec{\mathbf{u}}$ gives the same result as plugging in $\vec{\mathbf{u}}$ to the first column of a 3x3 matrix whose other two columns have the coordinates of $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, and computing the determinant. + +
+ +If you think in terms of duality and turn this matrix on its side, we're looking for a special 3d vector, that I'll call $\vec{\mathbf{p}}$, such that taking the dot product between $\vec{\mathbf{p}}$ and any other vector $\vec{\mathbf{u}}$ gives the same result as plugging in $\vec{\mathbf{u}}$ to the first column of a matrix and 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 $\vec{\mathbf{p}}$ and $\vec{\mathbf{u}}$ will give + +$$ +(\text{something}) \cdot x + (\text{something}) \cdot y + (\text{something}) \cdot z +$$ + +Those somethings are the coordinates of $\vec{\mathbf{p}}$. When you compute the determinant on the right, you can organize it to look like + +$$ +(v_2 w_3 - v_3 w_2) \cdot x + (v_3 w_1 - v_1 w_3) \cdot y + (v_1 w_2 - v_2 w_1) \cdot z +$$ + +This shows what those "somethings" are, and hence what the coordinates of the vector $\mathbf{p}$ are. + +
+ +So the answers to what those "somethings" are will give the coordinates of the vector $\vec{\mathbf{p}}$ that we're looking for. + +
+ +But this should feel very familiar to anyone who's actually worked through a cross product computation before: Collecting the constant terms that are multiplied by $x$, $y$ and $z$ like this is no different from plugging in the symbols $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$, and seeing which coefficients aggregate on each of these terms. It's just that plugging in $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$ is a way of signaling that we should interpret these three coefficients as coordinates of a vector. + +
+ +In other words, this funky computation can be thought of as the answer to a certain question: "What vector $\vec{\mathbf{p}}$ has the special property that when you take a dot product between $\vec{\mathbf{p}}$ and $\vec{\mathbf{u}}$, it gives the same result as plugging in $\vec{\mathbf{u}}$ to the first column of a matrix whose other two columns have the coordinates of $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, then computing the determinant?" + +
+ +Now for the cool part, which ties this all together with the geometric understanding of the cross product I showed earlier. Let's ask that same question again, but this time we're going to try to answer it geometrically, instead of computationally: "What 3d vector $\vec{\mathbf{p}}$ has the property that when you take a dot product between $\vec{\mathbf{p}}$ and some other vector $\vec{\mathbf{u}}$, it gives the same value as if you took the signed volume of the parallelepiped defined by this vector $\vec{\mathbf{u}}$ along with $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$?" + +
+ +Remember, the geometric interpretation of the dot product between a vector $\vec{\mathbf{p}}$ and some other vector is to project that other vector onto $\vec{\mathbf{p}}$, and multiply the length of the projection by the length of $\vec{\mathbf{p}}$. + +
+ +Here's one way to think of the volume of the parallelepided formed by $\vec{\mathbf{v}}$, $\vec{\mathbf{w}}$, and $\vec{\mathbf{u}}$: take the area of the parallelogram defined by $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, and multiply it not by the length of $\vec{\mathbf{u}}$, but by the component of $\vec{\mathbf{u}}$ which is perpendicular to that parallelogram. + +
+ +In other words, consider, the way our linear function works on a given vector is to project that vector onto the line perpendicular to both $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, then multiply the length of the projection by the area of the parallelogram spanned by $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. But this is the same thing as taking a dot product between $\vec{\mathbf{u}}$ and a vector perpendicular to $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ with a length equal to the area of the parallelogram! + +
+ +If you choose the appropriate direction for that vector, the cases where the dot product is negative will be the same as those when the right-hand-rule orientation of $\vec{\mathbf{u}}$, $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ is negative. For example, a quick way to reverse the orientation is to swap $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ which inverts the volume of the parallelepided. + +
+ +This means we just found a vector $\vec{\mathbf{p}}$ so that taking a dot product between $\vec{\mathbf{p}}$ and some vector $\vec{\mathbf{u}}$ is the same thing as computing the determinant of a 3x3 matrix whose columns contain the coordinates of that vector, $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$! + +
+ +So the answer we found computationally earlier using the notational trick must correspond to this vector. This is the fundamental reason why the computation and the geometric interpretation are related! + +## Conclusion + +To sum up what just happened, we defined a linear transformation from 3d space to the number line defined in terms of the vectors $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. + +
+ +Then we went through two different ways to think about the dual vector of this transformation, the vector such that applying the transformation is the same as taking a dot product with this vector. + +
+ +On the one hand, thinking about the dual vector from a computational approach leads us to the trick of plugging in the symbols $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$ to the first column of a matrix and computing the determinant. + +
+ +But thinking geometrically, we can deduce that this dual vector must be perpendicular to $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$, with a length equal to the area of the parallelogram spanned out by those two vectors. Also, it's direction will be determined by the right-hand rule. + +
+ +For both approaches, finding the dual vector of this linear transformation gives us a much deeper understanding of the result of the cross product between two vectors $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ than just the formula. + +
+ +Next up are two important concepts in linear algebra: Cramer's Rule and Change of basis. + +## Practice + +
+ + + +Using your right-hand point your forefinger first in the direction of $\vec{\mathbf{v}}$ and then your middle finger in the direction of $\vec{\mathbf{w}}$. This locks the orientation of your hand in space so the only direction your thumb can point is up. + +
+ +So the cross product of $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ must be pointing in the upward direction. + + + + +
+ + + + diff --git a/public/content/lessons/2016/cross-products-extended/notes.txt b/public/content/lessons/2016/cross-products-extended/notes.txt new file mode 100644 index 00000000..19f4ba99 --- /dev/null +++ b/public/content/lessons/2016/cross-products-extended/notes.txt @@ -0,0 +1,8 @@ +- The last line "Next up is a really important concept in linear algebra: Change of basis." contradicts that the next lesson is about Cramer's Rule. +- Should lesson links try and avoid mentioning the chapter number and instead use the chapter title or something like it? +- The typesetting of $\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$ disrupts the paragraph grouping, so I updated the text to use $\vec{\mathbf{u}}$ instead. I've also had it where teachers write vectors horizontally, but I think I prefer the variable approach instead? +- The moving camera really helps the brain understand the layout of 3d space, I experimented with drawing the corresponding rectangular prisms for the vectors to help the reader understand the geometry from a static perspective. This is the same approach that I would use if I was drawing the vectors by hand with some vanishing points. +- The parallelepiped whicht talks about volume is drawn with $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ drawn lying on the $xy$-plane... should this be pointed out to the student? +- TODO: animate figure that has the caption "Even though we are drawing the vector..." +- Potential question: What is the signed volume of this parallelepiped? +- Potential question: Computing the determinant with variables for u (x y z) and collecting like terms diff --git a/public/content/lessons/2016/cross-products/figures/determinant-basis-vectors-transformed-orientation.png b/public/content/lessons/2016/cross-products/figures/compute-with-determinant/determinant-basis-vectors-transformed-orientation.png similarity index 100% rename from public/content/lessons/2016/cross-products/figures/determinant-basis-vectors-transformed-orientation.png rename to 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a/public/content/lessons/2016/cross-products/index.mdx +++ b/public/content/lessons/2016/cross-products/index.mdx @@ -124,11 +124,9 @@ This is because a matrix whose columns represent $\vec{\mathbf{v}}$ and $\vec{\m The determinant is all about measuring how areas change due to a transformation, and the prototypical area to look at is the unit square resting on $\hat{\imath}$ and $\hat{\jmath}$. After the transformation, that square turns into the parallelogram we care about. So the determinant, which generally measures the factor by which areas are changed, gives the area of this parallelogram, since it evolved from a square that started with area 1.
- - What's more, if $\vec{\mathbf{w}}$ is a _clockwise_ rotation away from $\vec{\mathbf{v}}$, it means orientation is flipped during the transformation, which is what it means for a determinant to be negative.
-## 3d cross product, standard view +## Standard 3d View + +Now, even though everything we just did is a perfectly fine mathematical operation, what I just described is technically not the cross product. The true cross product is something that combines two 3d vectors to get a new 3d vector. -Now, even though everything we just did is a perfectly fine mathematical operation, what I just described is technically not the cross product. The true cross product is something that combines two 3d vectors to get a new 3d vector. You still consider the parallelogram defined by the two vectors, and the area of that parallelogram still plays a big role. To be concrete let's say that area is $2.5$ for the vectors shown here. +$$ +\left[\begin{array}{l} +v_1 \\ +v_2 \\ +v_3 +\end{array}\right] +\times +\left[\begin{array}{l} +w_1 \\ +w_2 \\ +w_3 +\end{array}\right] += +\left[\begin{array}{l} +p_1 \\ +p_2 \\ +p_3 +\end{array}\right] +$$ + +You still consider the parallelogram defined by the two vectors, and the area of that parallelogram still plays a big role. To be concrete the area is $9$ units for the vectors shown here.
-The 3d cross product does not return a number, it returns a vector. This new vector's length will be the area of that parallelogram, which in this example is 2.5, and the direction of that new vector is perpendicular +The 3d cross product does not return a number, it returns a vector. This new vector's length will be the area of that parallelogram, which in this example is $9$, and the direction of that new vector is perpendicular to that parallelogram.
- - -But which way? There are two possible vectors with length $2.5$ pointing perpendicular to a given plane. This is where the right-hand rule comes in. Point the forefinger of your right hand in the direction of $\vec{\mathbf{v}}$, and stick out your middle finger in the direction of $\vec{\mathbf{w}}$. Now when you point up your thumb, that's the direction of the cross product. +But which way? There are two possible vectors with length $9$ pointing perpendicular to a given plane. This is where the right-hand rule comes in. Point the forefinger of your right hand in the direction of $\vec{\mathbf{v}}$, and stick out your middle finger in the direction of $\vec{\mathbf{w}}$. Now when you point up your thumb, that's the direction of the cross product.
For example, let $\vec{\mathbf{v}}$ be the vector with length $2$ in the $z$ direction, and let $\vec{\mathbf{w}}$ be the vector with length $2$ in the $y$ direction. The parallelogram they define in this simple example is actually a square, since they're perpendicular, and the area of that square is $4$. Using the right hand rule, their cross product should point in the negative $x$ direction, so this cross product is $-4 \hat{\imath}$.
-## Computing + -There is a formula you could memorize for this cross product: +Using what we know about computing the cross product so far you can draw the parallelogram formed betwen $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$. + +
+ +Which, in this case, is a rectangle with an area of $6$ units. So we know the resulting vector $\vec{\mathbf{p}}$ will have a length of $6$ and will be parallel to the $xy$-coordinate plane, so it will lie either on the positive $z$-axis or the negative $z$-axis. Applying the right-hand rule tell us that it is the vector pointing in the negative direction on the $z$-axis.
-However, it's common and easier to remember a certain process involving the 3d determinant. +Writing this out gives us the following computation. + +$$ +\left[\begin{array}{c} 2 \\ 0 \\ 0 \end{array}\right] +\times +\left[\begin{array}{c} 0 \\ -3 \\ 0 \end{array}\right] += +\left[\begin{array}{c} 0 \\ 0 \\ -6 \end{array}\right] +$$ + + + +## Computing + +There is a formula you could memorize for this cross product: + +$$ +\left[\begin{array}{l} +v_1 \\ +v_2 \\ +v_3 +\end{array}\right] \times\left[\begin{array}{l} +w_1 \\ +w_2 \\ +w_3 +\end{array}\right]=\left[\begin{array}{l} +v_2 \cdot w_3-w_2 \cdot v_3 \\ +v_3 \cdot w_1-w_3 \cdot v_1 \\ +v_1 \cdot w_2-w_1 \cdot v_2 +\end{array}\right] +$$ -This process looks truly strange at first. Write a 3d matrix where the second column is the coordinates of $\vec{\mathbf{v}}$, and the third column is the coordinates $\vec{\mathbf{w}}$. Then for the first column, write in the basis vector $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$. Then, you compute the determinant of this matrix. +However, it's common and easier to remember a certain process involving the 3d determinant. This process looks truly strange at first. Write a 3d matrix where the second column is the coordinates of $\vec{\mathbf{v}}$, and the third column is the coordinates $\vec{\mathbf{w}}$. Then for the first column, write in the basis vector $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$. Then, you compute the determinant of this matrix.
The silliness is probably clear here. What on earth does it mean to put a vector in as the entry to a matrix? Students are often told that this is just a notational trick; when you carry out all the computations as if $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$ were numbers. Then you get some linear combination of those basis vectors, and the vector defined by that linear combination, students are told to just believe, is the unique vector perpendicular to $\vec{\mathbf{v}}$ and $\vec{\mathbf{w}}$ whose magnitude is the area of the appropriate parallelogram, and whose direction obeys the right-hand-rule.
And sure, in some sense this is a notational trick, but there is a reason for doing it. It's not just a coincidence that the determinant is once again important, and putting the basis vectors in these slots is not just random. To understand where this all comes from, it helps to use the idea of duality that I introduced in the last chapter. This concept is a bit heavy, though, so I'm putting it into a separate follow-on chapter for those of you curious to learn more. -Arguably, it falls outside the essence of linear algebra, so you can feel comfortable if you want to skip that next chapter. But for those of you willing to go a bit deeper, it's just really cool. +Arguably, it falls outside the essence of linear algebra, so you can feel comfortable if you want to skip that next chapter and go straight to the lesson on Cramer's Rule. But for those of you willing to go a bit deeper, it's just really cool. diff --git a/public/content/lessons/2016/cross-products/notes.txt b/public/content/lessons/2016/cross-products/notes.txt index a9c3a2ff..6cda1ae6 100644 --- a/public/content/lessons/2016/cross-products/notes.txt +++ b/public/content/lessons/2016/cross-products/notes.txt @@ -2,4 +2,5 @@ - TODO: change colors of "two dimensions" section to be yellow and purple - Possibly reword sentence containing the word "prototypical" - Should the parallelogram always be the same color as the first operand vector in the cross product? -- Should there be some comprehension questions applying the notational trick formula? Or should the student be encouraged into the next more in-depth lesson? \ No newline at end of file +- Should there be some comprehension questions applying the notational trick formula? Or should the student be encouraged into the next more in-depth lesson? +- TODO: replace `./figures/compute-with-determinant/determinant-basis-vectors-transformed-orientation.png` with image callback from chapter five \ No newline at end of file