Fall 2025
MATH 224: Multivariable Calculus and Geometry. I
Branko Ćurgus

Friday, October 3, 2025

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• Today we talked about Section 2.4 The Cross Product in OpenStax Calculus Volume 3. Do as many problems as you can from 183 to . In particular: 183, 184, 185, 191, 196, 197, 205, 207, 209, 214, 223, 224, 225, 227.
• Let \(\mathbf a=\langle a_1,a_2,a_3\rangle\) and \(\mathbf b=\langle b_1,b_2,b_3\rangle\). The coordinate definition of the cross product \(\mathbf a\times\mathbf b\) is \[ \mathbf a\times\mathbf b =\Big\langle a_2 b_3-a_3 b_2,\; -(a_1 b_3 - a_3 b_1),\; a_1 b_2-a_2 b_1 \Big\rangle. \] In the following items we will prove the following properties of the cross product:
  • For arbitrary vectors \(\mathbf a=\langle a_1,a_2,a_3\rangle\) and \(\mathbf b=\langle b_1,b_2,b_3\rangle\) we have \[ (\mathbf a\times\mathbf b)\cdot \mathbf a = 0 \] and \[ (\mathbf a\times\mathbf b)\cdot \mathbf b = 0. \]
  • For arbitrary vectors \(\mathbf a=\langle a_1,a_2,a_3\rangle\) and \(\mathbf b=\langle b_1,b_2,b_3\rangle\) we have \[ \mathbf b \times\mathbf a = - (\mathbf a\times\mathbf b) \]
  • For arbitrary vectors \(\mathbf a=\langle a_1,a_2,a_3\rangle\), \(\mathbf b=\langle b_1,b_2,b_3\rangle,\) and \(\mathbf c=\langle c_1,c_2,c_3\rangle\) and arbitrary scalars \(\alpha, \beta \in \mathbb{R}\) we have \[ (\alpha\mkern1.5mu \mathbf a+\beta\mkern1.5mu \mathbf b)\times \mathbf c = \alpha (\mathbf a\times \mathbf c) + \beta(\mathbf b\times \mathbf c), \] and \[ \mathbf a \times (\alpha\mkern1.5mu \mathbf b+\beta\mkern1.5mu \mathbf c) = \alpha (\mathbf a\times \mathbf b) + \beta(\mathbf a \times \mathbf c). \]
  • For arbitrary vectors \(\mathbf a=\langle a_1,a_2,a_3\rangle\) and \(\mathbf b=\langle b_1,b_2,b_3\rangle\) we have \[ \| \mathbf a \times\mathbf b \|^2 = \| \mathbf a \|^2 \| \mathbf b \|^2 - (\mathbf{a} \cdot \mathbf{b})^2. \]
  • For arbitrary vectors \(\mathbf a=\langle a_1,a_2,a_3\rangle\) and \(\mathbf b=\langle b_1,b_2,b_3\rangle\) we have \[ \|\mathbf a\times \mathbf b\| = \|\mathbf a\| \|\mathbf b\| \mkern 2mu \sin\theta, \] where the expression \[ \|\mathbf a\| \|\mathbf b\| \mkern 2mu \sin\theta \] gives the area of the parallelogram formed by the vectors \(\mathbf a\) and \(\mathbf b.\)
• In this item we prove that the cross product \(\mathbf a\times\mathbf b\) is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\) by calculating the dot products \begin{align*} (\mathbf a\times\mathbf b)\cdot \mathbf a & = (a_2 b_3-a_3 b_2 )a_1 - (a_1 b_3 - a_3 b_1) a_2 + (a_1 b_2-a_2 b_1 )a_3 \\[2pt] & = a_1 a_2 b_3 - a_1 a_3 b_2 - a_1 a_2 b_3 + a_2 a_3 b_1 + a_1 a_3 b_2 - a_2 a_3 b_1 \\[2pt] &=\underbrace{a_1 a_2 b_3 - a_1 a_2 b_3}_{0} \;+\;\underbrace{a_1 a_3 b_2 - a_1 a_3 b_2}_{0} \;+\;\underbrace{a_2 a_3 b_1 - a_2 a_3 b_1}_{0} \\[2pt] &=0. \end{align*} and \begin{align*} (\mathbf a\times\mathbf b)\cdot \mathbf b &= (a_2 b_3-a_3 b_2 )b_1 -(a_1 b_3 - a_3 b_1) b_2 + (a_1 b_2-a_2 b_1 )b_3 \\[2pt] &= a_2 b_1 b_3 - a_3 b_1 b_2 - a_1 b_2 b_3 + a_3 b_1 b_2 + a_1 b_2 b_3 - a_2 b_1 b_3 \\[2pt] &= \underbrace{a_2 b_1 b_3 - a_2 b_1 b_3}_{0} \;+\;\underbrace{a_3 b_1 b_2 - a_3 b_1 b_2}_{0} \;+\;\underbrace{a_1 b_2 b_3 - a_1 b_2 b_3}_{0} \\[2pt] &=0. \end{align*}
• In the next item we will prove the following identity: \[ \| \mathbf a \times\mathbf b \|^2 = \| \mathbf a \|^2 \| \mathbf b \|^2 - (\mathbf{a} \cdot \mathbf{b})^2. \] It is a long algebraic calculation. A human could do this by hand, and humans use to do this by hand in prior centuries. But in the twentyfirst century, I aksed ChatGPT to do it for us. I present what ChatGPT wrote below.
• Let \(\mathbf a=\langle a_1,a_2,a_3\rangle\), \(\mathbf b=\langle b_1,b_2,b_3\rangle\). Recall \[ \mathbf a\times \mathbf b = \bigl\langle a_2 b_3 - a_3 b_2,\; -(a_1 b_3 - a_3 b_1),\; a_1 b_2 - a_2 b_1 \bigr\rangle, \] and calculate \begin{align*} \|\mathbf a\times \mathbf b\|^2 &= (a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2 \\[6pt] &= \bigl(a_2^2 b_3^2 - 2 a_2 a_3 b_2 b_3 + a_3^2 b_2^2\bigr) \\[2pt] &\mkern 50mu + \bigl(a_3^2 b_1^2 - 2 a_1 a_3 b_1 b_3 + a_1^2 b_3^2\bigr) \\[2pt] &\mkern 70mu + \bigl(a_1^2 b_2^2 - 2 a_1 a_2 b_1 b_2 + a_2^2 b_1^2\bigr) \\[6pt] &= \underbrace{\bigl(a_1^2 b_2^2 + a_2^2 b_3^2 + a_3^2 b_1^2\bigr) }_{Sb_{231}} + \underbrace{\bigl( a_1^2 b_3^2 + a_2^2 b_1^2 + a_3^2 b_2^2 \bigr)}_{Sb_{312}} \\[2pt] &\mkern 40mu - 2 \underbrace{\bigl( a_1 a_2 b_1 b_2 + a_1 a_3 b_1 b_3 + a_2 a_3 b_2 b_3 \bigr)}_{S4} \\[6pt] & = Sb_{231} + Sb_{312} -2 \mkern 2mu S4 \end{align*} Now compare with \begin{align*} \|\mathbf a\|^2 \|\mathbf b\|^2 &= \bigl(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2 \bigr) \\[6pt] &= a_1^2 b_1^2 + a_1^2 b_2^2 + a_1^2 b_3^2 \\ &\mkern 50mu + a_2^2 b_1^2 + a_2^2 b_2^2 + a_2^2 b_3^2 \\ &\mkern 70mu + a_3^2 b_1^2 + a_3^2 b_2^2 + a_3^2 b_3^2 \\[6pt] &= \underbrace{\bigl( a_1^2 b_1^2 + a_2^2 b_2^2 + a_3^2 b_3^2 \bigr)}_{Sb_{123}} \\[2pt] & \mkern 40mu + \underbrace{\bigl( a_1^2 b_2^2 + a_2^2 b_3^2 + a_3^2 b_1^2 \bigr)}_{Sb_{231}} \\[2pt] & \mkern 60mu + \underbrace{\bigl( a_1^2 b_3^2 + a_2^2 b_1^2 + a_3^2 b_2^2 \bigr)}_{Sb_{312}} \\[6pt] & = Sb_{123} + Sb_{231} + Sb_{312} \end{align*} Also \begin{align*} (\mathbf a\cdot \mathbf b)^2 &= \bigl( a_1 b_1 + a_2 b_2 + a_3 b_3 \bigr)^2 \\[2pt] &= \underbrace{\bigl( a_1^2 b_1^2 + a_2^2 b_2^2 + a_3^2 b_3^2 \bigr)}_{Sb_{123}} + 2 \underbrace{\bigl( a_1 a_2 b_1 b_2 + a_1 a_3 b_1 b_3 + a_2 a_3 b_2 b_3 \bigr)}_{S4} \\[2pt] & = Sb_{123} + 2 \mkern 2mu S4. \end{align*} Therefore \begin{align*} \|\mathbf a\|^2 \|\mathbf b\|^2 - (\mathbf a\cdot\mathbf b)^2 & = \bigl(Sb_{123} + Sb_{231} + Sb_{312}\bigr) - \bigl( Sb_{123} + 2 \mkern 2mu S4 \bigr) \\[2pt] & = Sb_{231} + Sb_{312} - 2 \mkern 2mu S4 \\[2pt] & = \|\mathbf a\times \mathbf b\|^2. \end{align*} Thus the expressions for \(\|\mathbf a\times \mathbf b\|^2\) and \(\|\mathbf a\|^2 \|\mathbf b\|^2 - (\mathbf a\cdot\mathbf b)^2\) are identical. That is we have proved \[ \boxed{\;\|\mathbf a\times \mathbf b\|^2 = \|\mathbf a\|^2\|\mathbf b\|^2 - (\mathbf a\cdot \mathbf b)^2\;} \]
• Why is \[ \boxed{\;\|\mathbf a\times \mathbf b\|^2 = \|\mathbf a\|^2\|\mathbf b\|^2 - (\mathbf a\cdot \mathbf b)^2\;} \] important? Let us obtain the geometric expression for the right-hand side: \begin{align*} \|\mathbf a\|^2\|\mathbf b\|^2 - (\mathbf a\cdot \mathbf b)^2 & = \|\mathbf a\|^2\|\mathbf b\|^2 - \bigl(\|\mathbf a \| \|\mathbf b\| \mkern 2mu \cos\theta \bigr)^2 \\[2pt] & = \|\mathbf a\|^2\|\mathbf b\|^2 - \|\mathbf a\|^2\|\mathbf b\|^2 (\cos\theta )^2 \\[2pt] & = \|\mathbf a\|^2\|\mathbf b\|^2 \bigl( 1 - (\cos\theta )^2 \bigr) \\[2pt] & = \|\mathbf a\|^2\|\mathbf b\|^2 (\sin\theta )^2 \end{align*} Therefore \[ \boxed{\ \|\mathbf a\times \mathbf b\|^2 =\|\mathbf a\|^2\|\mathbf b\|^2 (\sin\theta )^2 \ } \] Recall that \(\theta\) is the between \(\mathbf a\) and \(\mathbf b\) such that \(0 \leq \theta \leq \pi.\) Therefore \(0 \leq \sin\theta \leq 1,\) that is \(\sin\theta\) is nonnegative number. The numbers \(\|\mathbf a\times \mathbf b\|,\) \(\|\mathbf a\|\) and \(\|\mathbf b\|\) are positive. Therefore \[ \boxed{\ \|\mathbf a\times \mathbf b\| = \|\mathbf a\| \|\mathbf b\| \mkern 2mu \sin\theta \ } \] The expression \[ \boxed{\ \|\mathbf a\| \|\mathbf b\| \mkern 2mu \sin\theta \ } \] gives the area of the parallelogram formed by the vectors \(\mathbf a\) and \(\mathbf b.\) This is proved in Theorem 2.8 on page 157 in the OpenStax textbook.
• In conclusion, by establishing the equality \[ \boxed{\ \|\mathbf a\times \mathbf b\| = \|\mathbf a\| \|\mathbf b\| \mkern 2mu \sin\theta \ } \] we have proved that the magnitude of the cross product \(\mathbf a\times \mathbf b,\) as given by its coordinate definition, is precisely the area formed by the vectors \(\mathbf a\) and \(\mathbf b.\)
• Let \(\mathbf a = \langle a_1, a_2, a_3\rangle,\) and \(\mathbf b = \langle b_1, b_2, b_3 \rangle,\). By definition, \[ \mathbf u\times\mathbf v =\bigl\langle u_2 v_3 - u_3 v_2,\; -(u_1 v_3 - u_3 v_1),\; u_1 v_2 - u_2 v_1 \bigr\rangle. \] Hence, \begin{align*} \mathbf a\times\mathbf b &=\bigl\langle a_2 b_3 - a_3 b_2,\; -(a_1 b_3 - a_3 b_1),\; a_1 b_2 - a_2 b_1 \bigr\rangle, \end{align*} and \begin{align*} \mathbf b\times\mathbf a &=\bigl\langle b_2 a_3 - b_3 a_2,\; -(b_1 a_3 - b_3 a_1),\; b_1 a_2 - b_2 a_1 \bigr\rangle \\[4pt] &=\bigl\langle -(a_2 b_3 - a_3 b_2),\; -(-a_1 b_3 + a_3 b_1),\; -(a_1 b_2 - a_2 b_1) \bigr\rangle \\[4pt] &=-\,\bigl\langle a_2 b_3 - a_3 b_2,\; -(a_1 b_3 - a_3 b_1),\; a_1 b_2 - a_2 b_1 \bigr\rangle. \end{align*} Therefore, \[ \boxed{\;\mathbf b\times\mathbf a = -(\mathbf a\times\mathbf b).\;} \]
• Let \(\mathbf a = \langle a_1, a_2, a_3\rangle\), \(\mathbf b = \langle b_1, b_2, b_3 \rangle,\) \(\mathbf c=\langle c_1,c_2,c_3\rangle,\) and \(\alpha,\beta\in\mathbb{R}.\) Recall \[ \mathbf u\times\mathbf v =\bigl\langle u_2 v_3-u_3 v_2,\; -(u_1 v_3 - u_3 v_1), \; u_1 v_2-u_2 v_1 \bigr\rangle. \] We compute \((\alpha\mkern1.5mu \mathbf a+\beta\mkern1.5mu \mathbf b)\times \mathbf c\) and show it equals \(\alpha (\mathbf a\times \mathbf c)+\beta(\mathbf b\times \mathbf c).\) Recall that \[ \alpha\mkern1.5mu \mathbf a+\beta\mkern1.5mu\mathbf b = \bigl\langle \alpha\mkern1.5mu a_1 + \beta\mkern1.5mu b_1,\;\alpha\mkern1.5mu a_2 + \beta\mkern1.5mu b_2,\; \alpha\mkern1.5mu a_3 + \beta\mkern1.5mu b_3 \bigr\rangle. \] We calculate \begin{align*} (\alpha\mkern1.5mu \mathbf a + \beta\mkern1.5mu \mathbf b) \times \mathbf c & = \Big\langle (\alpha\mkern1.5mu a_2 + \beta\mkern1.5mu b_2) c_3 - (\alpha\mkern1.5mu a_3 + \beta\mkern1.5mu b_3)c_2,\; \\ &\hspace{1.3cm} -\bigl((\alpha\mkern1.5mu a_1 + \beta\mkern1.5mu b_1) c_3 - (\alpha\mkern1.5mu a_3 + \beta\mkern1.5mu b_3)c_1 \bigr),\\ &\hspace{2.6cm} (\alpha\mkern1.5mu a_1 + \beta\mkern1.5mu b_1)c_2 - (\alpha\mkern1.5mu a_2 + \beta\mkern1.5mu b_2) c_1 \Big\rangle \\[2pt] &= \Big\langle \alpha\mkern1.5mu (a_2 c_3-a_3 c_2) + \beta\mkern1.5mu (b_2 c_3-b_3 c_2),\; \\ &\hspace{1.3cm} -\alpha\mkern1.5mu (a_1 c_3 - a_3 c_1) - \beta\mkern1.5mu (b_1 c_3 - b_3 c_1),\\ &\hspace{2.6cm} \alpha\mkern1.5mu (a_1 c_2-a_2 c_1) + \beta\mkern1.5mu (b_1 c_2-b_2 c_1) \Big\rangle \\[2pt] &=\Big\langle \alpha\mkern1.5mu (a_2 c_3-a_3 c_2),\; -\alpha\mkern1.5mu (a_1 c_3 - a_3 c_1),\; \alpha\mkern1.5mu (a_1 c_2-a_2 c_1) \Big\rangle \\ &\hspace{1.3cm} + \Big\langle \beta\mkern1.5mu (b_2 c_3-b_3 c_2),\; - \beta\mkern1.5mu (b_1 c_3 - b_3 c_1), \; \beta\mkern1.5mu (b_1 c_2-b_2 c_1) \Big\rangle \\[2pt] & = \alpha\,\underbrace{\Big\langle a_2 c_3-a_3 c_2,\; -(a_1 c_3 - a_3 c_1),\; a_1 c_2-a_2 c_1 \Big\rangle}_{\mathbf a\times \mathbf c} \\ &\hspace{1.3cm} \;+\; \beta\,\underbrace{\Big\langle b_2 c_3-b_3 c_2,\; -(b_1 c_3 - b_3 c_1),\; b_1 c_2-b_2 c_1 \Big\rangle}_{\mathbf b\times \mathbf c}. \end{align*} Therefore, \[ \boxed{\ (\alpha\mkern1.5mu \mathbf a+\beta\mkern1.5mu \mathbf b)\times \mathbf c =\alpha\mkern1.5mu (\mathbf a\times \mathbf c)+\beta\mkern1.5mu (\mathbf b\times \mathbf c)\ }. \]

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Thursday, October 2, 2025

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• Given two nonzero vectors \(\mathbf{a}\) and \(\mathbf{b},\) what is the angle formed by these vectors? By the angle formed by the vectors \(\mathbf{a}\) and \(\mathbf{b}\), I mean an angle between \(0\) and \(\pi\) including \(0\) and \(\pi.\) To answer this question, we use the dot product, and this is one of the prominent usages of the dot product.

  From the geometric definition of the dot product we have \[ \mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \mkern 1mu \| \mathbf{b} \| \mkern 1mu \cos\theta, \] where \(\theta\) is the angle angle formed by \(\mathbf{a}\) and \(\mathbf{b}.\) From this formula we calculate \[ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{ \| \mathbf{a} \| \mkern 1mu \| \mathbf{b} \|}. \] Therefore, \[ \theta = \arccos \left(\frac{\mathbf{a} \cdot \mathbf{b}}{ \| \mathbf{a} \| \mkern 1mu \| \mathbf{b} \|} \right). \]

  Inspired by the the fact that we used the arccos function above, I briefly reviewed the concept of a function. I wrote more about Functions on this webpage: Functions

• Today I proved Theorem of Thales using vectors. The other two theorems are exercises in the textbook. I added vectors to make it easier for you to write your own proofs. On Tuesday I posted two more theorems from Euclidean Geometry which can be proved using vectors. Understanding these proofs is a good practice with vectors and mathematical reasoning.

• Theorem (a characterization of a rectangle) (often used in carpentry).

  Let \(ABCD\) be a parallelogram with vertices \(A, B, C, D\) and the diagonals \(\overline{AC}\) and \(\overline{BD}\), as pictured below. Then \(ABCD\) is a rectangle if and only if the diagonals \(\overline{AC}\) and \(\overline{BD}\) have the same length.

  Proof (using vectors).





• Theorem (a characterization of a rhombus)

  Let \(ABCD\) be a parallelogram with vertices \(A, B, C, D\) and the diagonals \(\overline{AC}\) and \(\overline{BD}\), as pictured above. Then \(ABCD\) is a rhombus if and only if the diagonals \(\overline{AC}\) and \(\overline{BD}\) perpendicular.

  Proof (using vectors).

• Theorem of Thales. If \(A\), \(B\), and \(C\) are points on a circle with diameter \(\overline{AC}\), then the angle at \(B\) is a right angle. Conversely, if the triangle \(ABC\) is a right triangle with the right angle at \(B\), then \(B\) lies on the circle having \(\overline{AC}\) as its diameter.

  

  

  Proof (using vectors). Assume that \(A\), \(B\), and \(C\) are points on a circle with diameter \(\overline{AC}.\) Our goal is to prove that the line segments \(\overline{BA}\) and \(\overline{BC}\) are orthogonal. The dot product of vectors is an efficient tool for proving orthogonality. Therefore we use vectors in this proof.

  Since \(\overline{AC}\) is a diameter of a circle, this circle's center is the midpoint of \(\overline{AC}.\) Denote this midpoint by \(O.\) Introduce the following vectors by their representative oriented line segments: \[ \mathbf{a}=\overrightarrow{BA}, \quad \mathbf{b}=\overrightarrow{BC}, \quad \mathbf{c}=\overrightarrow{OC}, \quad \text{and} \quad \mathbf{d}=\overrightarrow{BO}. \] Since \(O\) is the midpoint of \(\overline{AC},\) we have that the vector \(-\mathbf{c}\) is represented by \(\overrightarrow{OA}.\) The following algebraic relationships hold among the vectors that we introduced: \begin{align*} \mathbf{a} & = \mathbf{d} - \mathbf{c} \\ \mathbf{b} & = \mathbf{d} + \mathbf{c}. \end{align*} Since \(O\) is the center of a circle and the points (A\), \(B\), and \(C\) are on that circle, we have \[ \| \mathbf{c} \| = \| \mathbf{d} \|. \] Our goal is to prove that the line segments \(\overline{BA}\) and \(\overline{BC}\) are orthogonal. To prove this, we calculate \begin{align*} \mathbf{a} \cdot \mathbf{b} & = (\mathbf{d} - \mathbf{c})\cdot (\mathbf{d} + \mathbf{c}) \\ & = \mathbf{d} \cdot \mathbf{d} - \mathbf{c} \cdot \mathbf{d} + \mathbf{d} \cdot \mathbf{c} - \mathbf{c} \cdot \mathbf{c} \\ & = \mathbf{d} \cdot \mathbf{d} - \mathbf{d} \cdot \mathbf{c} + \mathbf{d} \cdot \mathbf{c} - \mathbf{c} \cdot \mathbf{c} \\ & = \mathbf{d} \cdot \mathbf{d} - \mathbf{c} \cdot \mathbf{c} \\ & = \| \mathbf{d} \|^2 - \| \mathbf{c} \|^2 \\ & = 0. \end{align*} Consequently, the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are orthogonal. Therefore, the line segments \(\overline{BA}\) and \(\overline{BC}\) are orthogonal. QED

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Wednesday, October 1, 2025

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• Today we finished Section 2.3 Dot Product in OpenStax Calculus Volume 3. Do as many problems as you can from 123 to 172. In particular: 124, 125, 127, 130, 131, 135, 137, 138, 141, 142, 143, 145, 147, 148, 150, 151, 153, 154, 155, 157, 158, 159, 161, 165, 168, 170, 172.
• Again, I emphasize that to understand the dot product it is essential to review the Law of Cosine that is taught in trigonometry classes. Here is my webpage with a proof of the Law of Cosines and its connection to dot product.

• An important application of dot product is the calculation of the orthogonal projection.

  Here is my presentation of the orthogonal projection of \(\mathbf{v}\) onto \(\mathbf{u}.\) Given two nonzero vectors \(\mathbf{u}\) and \(\mathbf{v}\), the orthogonal projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is a special scaling of \(\mathbf{u}\), call it \[ \alpha \mkern1mu \mathbf{u}, \qquad \alpha \in \mathbb{R}, \] which has a very special property that the difference \[ \mathbf{v} - \alpha \mkern1mu \mathbf{u} \qquad \text{is orthogonal to} \qquad \mathbf{u}. \] Using the orthogonality stated on the line above, we can calculate \(\alpha\) in terms of \(\mathbf{u}\) and \(\mathbf{v}\): \[ \bigl( \mathbf{v} - \alpha \mkern1mu\mathbf{u}\bigr) \cdot \mathbf{u} = 0. \] Now we use the distributivity property of the dot product to get \[ \mathbf{v} \cdot \mathbf{u} - \alpha \mkern1mu (\mathbf{u} \cdot \mathbf{u}) = 0. \] The last equality we can solve for \(\alpha\): \[ \alpha = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}}. \] Now we have the formula for the projection vector of \(\mathbf{v}\) onto \(\mathbf{u}\): \[ \operatorname{proj}_{\mathbf{u}} (\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}}\ \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2} \ \mathbf{u} \]

  Now I can write a formula in which we see what is so special about the projection vector of \(\mathbf{v}\) onto \(\mathbf{u}\): \[ \mathbf{v} = \underbrace{\frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2}\,\mathbf{u}}_{\text{a multiple of } \mathbf{u}} + \underbrace{ \left(\mathbf{v} - \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2}\,\mathbf{u} \right)}_{\text{orthogonal to } \mathbf{u}} \]

• Orthogonal projections \(\operatorname{proj}_{\mathbf{u}}(\mathbf{v})\) and related vectors in color

  

  

  
  
  

  The animation below illustrates the fact that the projection \(\operatorname{proj}_{\mathbf{u}}(\mathbf{v})\) does not change when \(\mathbf{u}\) is replaced by its nonzero scalar multiple.

  

• The Definition of the projection vector on page 140 is confused and formula (2.7) is wrong. Below is a snippet from the definition. Reading just the orange words, one does not know whether the projection is in the same direction as \(\mathbf{u}\), or it can also be in the opposite direction of \(\mathbf{u}.\) In fact, if the angle between \(\mathbf{v}\) and \(\mathbf{u}\) is acute, then the projection of \(\mathbf{v}\) onto \(\mathbf{u}\) points in the same direction as \(\mathbf{u}\), and if the angle between \(\mathbf{v}\) and \(\mathbf{u}\) is obtuse, then the projection of \(\mathbf{v}\) onto \(\mathbf{u}\) points in the opposite direction of \(\mathbf{u}.\) The projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is always colinear with \(\mathbf{u}.\) That is, it is a scalar multiple of \(\mathbf{u}.\)

  Below is the snippet of (2.7) on page 140: This is quite confusing. When you look at (2.7), on the left-hand side is a vector, while on the right-hand side is a scalar. They cannot be equal. Vector can never equal a scalar. I was not familiar with the concept scalar projection of \(\mathbf{v}\) onto \(\mathbf{u}.\) Based on my research, the common meaning of the scalar projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is not the length of the projection, but something a little more complicated. One could say that it is the signed length; it is the length when the projection and \(\mathbf{u}\) point in the same direction and it is the opposite of the length when the projection and \(\mathbf{u}\) point in different directions. Or it is the scalar that we use to scale the unit vector of \(\mathbf{u}\) to get the projection.

  So, the book is wrong for using the word "length," in the sentence with (2.7), but the formula for \(\operatorname{comp}_{\mathbf{u}} \mathbf{v}\) is correct when one removes \(\operatorname{proj}_{\mathbf{u}} \mathbf{v}.\)

  Here is an explanation of \(\operatorname{comp}_{\mathbf{u}} \mathbf{v}\), starting from the formula for \(\operatorname{proj}_{\mathbf{u}} \mathbf{v}\): \[ \operatorname{proj}_{\mathbf{u}}\mkern-5mu \mathbf{v} = \frac{\mathbf{v}\cdot\mathbf{u}}{\|\mathbf{u}\|^2}\,\mathbf{u} = \frac{\mathbf{v}\cdot\mathbf{u}}{\|\mathbf{u}\|}\,\left(\frac{1}{\|\mathbf{u}\|} \mathbf{u}\right) = \underbrace{ \frac{\mathbf{v}\cdot\mathbf{u}}{\|\mathbf{u}\|}}_{ \operatorname{comp}_{\mathbf{u}}\mkern-5mu\mathbf{v}} \mkern 7mu \underbrace{\left(\frac{1}{\|\mathbf{u}\|} \mathbf{u}\right)}_{\begin{array}{c} \text{the unit } \\ \text{vector of} \ \mathbf{u} \end{array}}. \]

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Tuesday, September 30, 2025

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• Theorem about the diagonals in a deltoid. (A common name for a deltoid is a kite.)

  If \(ABCD\) is a deltoid with vertices \(A, B, C, D\), as shown below, then its diagonals \(\overline{AC}\) and \(\overline{BD}\) are perpendicular.

  Definition of a deltoid. A quadrilateral \(ABCD\) is called a deltoid if and only if it has two pairs of adjacent sides of equal length: \(|\overline{AB}| = |\overline{AD}|\) and \(|\overline{CB}| = |\overline{CD}|\).

  

  

  Proof (using vectors). Step 1. Introduce the following vectors by their representative line segments: \(\mathbf{a} = \overrightarrow{AD}\), \(\mathbf{b} = \overrightarrow{AB}\), \(\mathbf{c} = \overrightarrow{CB}\), and \(\mathbf{d} = \overrightarrow{CD}.\) Assume that \(|\overline{AB}| = |\overline{AD}|\) and \(|\overline{CB}| = |\overline{CD}|.\) With the vectors introduced, we assume that \[ \|\mathbf{a}\| = \|\mathbf{b}\| \quad \text{and} \quad \|\mathbf{c}\| = \|\mathbf{d}\|. \] Since \[ \mathbf{a} - \mathbf{d} + \mathbf{c} - \mathbf{b} = \mathbf{0}, \] we have that \[ \mathbf{a} - \mathbf{b} = \mathbf{d} - \mathbf{c}, \] and \[ \mathbf{a} - \mathbf{d} = \mathbf{b} - \mathbf{c}. \] The oriented line segment \(\overrightarrow{AC}\) is on one diagonal and it represents the vector \(\mathbf{a} - \mathbf{d} = \mathbf{b} - \mathbf{c}.\) The oriented line segment \(\overrightarrow{BD}\) is on the other diagonal and it represents the vector \(\mathbf{a} - \mathbf{b} = \mathbf{d} - \mathbf{c}.\) Next we will calculate the dot product \[ (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{d}). \]

  Step 2. Before calculating this dot product, we establish the following useful equality: \begin{align*} \mathbf{a} - \mathbf{b} & = \frac{1}{2} (\mathbf{a} - \mathbf{b}) + \frac{1}{2} (\mathbf{a} - \mathbf{b}) \\ & = \frac{1}{2} (\mathbf{a} - \mathbf{b}) + \frac{1}{2} (\mathbf{d} - \mathbf{c}) \\ & = \frac{1}{2} (\mathbf{a} + \mathbf{d}) - \frac{1}{2} (\mathbf{b} + \mathbf{c}). \end{align*} Hence we have \[ \mathbf{a} - \mathbf{b} = \frac{1}{2} (\mathbf{a} + \mathbf{d}) - \frac{1}{2} (\mathbf{b} + \mathbf{c}). \]

  Step 3. We use this equality to calculate \begin{align*} (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{d}) & = \left(\frac{1}{2} (\mathbf{a} + \mathbf{d}) - \frac{1}{2} (\mathbf{b} + \mathbf{c}) \right) \cdot (\mathbf{a} - \mathbf{d}) \\ & = \frac{1}{2} (\mathbf{a} + \mathbf{d}) \cdot (\mathbf{a} - \mathbf{d}) - \frac{1}{2} (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} - \mathbf{d}) \\ & = \frac{1}{2} \bigl( \|\mathbf{a}\|^2 - \|\mathbf{d}\|^2 \bigr) - \frac{1}{2} (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} - \mathbf{c}) \\ & = \frac{1}{2} \bigl( \|\mathbf{a}\|^2 - \|\mathbf{d}\|^2 \bigr) - \bigl( \|\mathbf{b}\|^2 - \|\mathbf{c}\|^2 \bigr) \\ & = \frac{1}{2} \bigl( \|\mathbf{a}\|^2 - \|\mathbf{b}\|^2 \bigr) + \frac{1}{2} \bigl( \|\mathbf{c}\|^2 - \|\mathbf{d}\|^2 \bigr) \\ \end{align*} Since we assume \[ \|\mathbf{a}\| = \|\mathbf{b}\| \quad \text{and} \quad \|\mathbf{c}\| = \|\mathbf{d}\|, \] we have \[ (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{d}) = \frac{1}{2} \bigl( \|\mathbf{a}\|^2 - \|\mathbf{b}\|^2 \bigr) + \frac{1}{2} \bigl( \|\mathbf{c}\|^2 - \|\mathbf{d}\|^2 \bigr) = 0. \] Thus, the vectors \(\mathbf{a} - \mathbf{b}\) and \(\mathbf{a} - \mathbf{d}\) are orthogonal. Therefore, the diagonals are orthogonal. QED.

• Varignon's Theorem

  Let \(ABCD\) be a quadrilateral with vertices \(A, B, C, D\) and denote by \(A'\) the midpoint of \(\overline{AB}\), by \(B'\) the midpoint of \(\overline{BC}\), by \(C'\) the midpoint of \(\overline{CD}\), and by \(D'\) the midpoint of \(\overline{DA}\), as pictured below. Then \(A'B'C'D'\) is a parallelogram.

  

  

  Proof (using vectors). Step 1. Introduce the following vectors by their representative line segments: \(\mathbf{a} = \overrightarrow{AB}\), \(\mathbf{b} = \overrightarrow{BC}\), \(\mathbf{c} = \overrightarrow{CD}\), and \(\mathbf{d} = \overrightarrow{DA}.\) Since \[ \mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}, \] we have that \[ \mathbf{b} + \mathbf{c} = - (\mathbf{a} + \mathbf{d}), \] (this vector is represented by the oriented line segment \(\overrightarrow{BD}\)) and \[ \mathbf{a} + \mathbf{b} = -(\mathbf{c} + \mathbf{d}). \] (this vector is represented by the oriented line segment \(\overrightarrow{AC}\)).

  Step 2. In triangle \(AA'D'\) we have that the oriented line segment \(\overrightarrow{A'D'}\) represents the vector \(-\frac{1}{2}(\mathbf{a} + \mathbf{d}).\) In triangle \(CB'C'\) we have that the oriented line segment \(\overrightarrow{B'C'}\) represents the vector \(\frac{1}{2}(\mathbf{b} + \mathbf{c}).\) Since we have \[ -\frac{1}{2}(\mathbf{a} + \mathbf{d}) = \frac{1}{2}(\mathbf{b} + \mathbf{c}), \] the oriented line segments \(\overrightarrow{A'D'}\) and \(\overrightarrow{B'C'}\) are equivalent. That is they are parallel and have equal length.

  Step 3. In triangle \(BA'B'\) we have that the oriented line segment \(\overrightarrow{A'B'}\) represents the vector \(\frac{1}{2}(\mathbf{a} + \mathbf{b}).\) In triangle \(DD'C'\) we have that the oriented line segment \(\overrightarrow{D'C'}\) represents the vector \(-\frac{1}{2}(\mathbf{d} + \mathbf{c}).\) Since we have \[ -\frac{1}{2}(\mathbf{c} + \mathbf{d}) = \frac{1}{2}(\mathbf{a} + \mathbf{b}), \] the oriented line segments \(\overrightarrow{A'B'}\) and \(\overrightarrow{D'C'}\) are equivalent. That is they are parallel and have equal length.

  Conclusion. We proved that the line segments \(\overline{A'D'}\) and \(\overline{B'C'}\) are parallel and have equal length and that the line segments \(\overline{A'B'}\) and \(\overline{D'C'}\) are parallel and have equal length. Therefore, \(A'B'C'D'\) is a parallelogram. QED.

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Monday, September 29, 2025

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• We finished Section 2.2 Vectors in Three Dimensions today.
• Tomorrow we will do Section 2.3 Dot Product in OpenStax Calculus Volume 3. Do as many problems as you can from 123 to 172. In particular: 124, 125, 127, 130, 131, 135, 137, 138, 141, 142, 143, 145, 147, 148, 150, 151, 153, 154, 155, 157, 158, 159, 161, 165, 168, 170, 172.
• To understand the dot product it is very useful to review the Law of Cosine that is taught in trigonometry classes. Since it is very important, I wrote a webpage for your convenience. Here is a proof of the Law of Cosines and its connection to dot product.

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Saturday, September 27, 2025

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• On Friday we started Section 2.2 Vectors in Three Dimensions in OpenStax Calculus Volume 3. Do as many problems as you can from 61 to 109. In particular: 61-68, 70, 73, 76, 79, 81, 86, 88, 91, 93, 94, 100, 104, 107-109.

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Thursday, September 25, 2025

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• A compass rose is a common tool for communicating direction in navigation. The compass rose below shows 32 distinct directions. Here are the names of all 32 directions:

  E - East	EbN - East-by-North	ENE - East-Northeast	NEbE - Northeast-by-East
  NE - Northeast	NEbN - Northeast-by-North	NNE - North-Northeast	NbE - North-by-East
  N - North	NbW - North-by-West	NNW - North-Northwest	NWbN - Northwest-by-North
  NW - Northwest	NWbW - Northwest-by-West	WNW - West-Northwest	WbN - West-by-North
  W - West	WbS - West-by-South	WSW - West-Southwest	SWbW - Southwest-by-West
  SW - Southwest	SWbS - Southwest-by-South	SSW - South-Southwest	SbW - South-by-West
  S - South	SbE - South-by-East	SSE - South-Southeast	SEbS - Southeast-by-South
  SE - Southeast	SEbE - Southeast-by-East	ESE - East-Southeast	EbS - East-by-South

  

• The role played by the compass rose in navigation, when we deal with vectors, is played by the unit circle in a coordinate plane.

  Below, I show the unit circle with 32 marked angles and their corresponding cosine and sine values. Note, however, that the angles on the compass rose are not the same as those on the unit circle: the compass rose uses angles that are multiples of \(\pi/16\) (11.25 degrees).

  For example, if we are given a vector \(\bigl\langle -1, \sqrt{3} \bigr\rangle\), we calculate the corresponding unit vector in the same direction: \(\bigl\langle -1/2, \sqrt{3}/2 \bigr\rangle\); then calculate the angle \(\theta\) such that \(\cos\theta = -1/2, \sin\theta = \sqrt{3}/2\); that is the angle \(\theta = 2 \pi /3\) and we state that the direction of the vector \(\bigl\langle -1, \sqrt{3} \bigr\rangle\) makes the angle of \(2 \pi /3\) with the positive direction of the horizontal axes.

  Although it is not done in mathematics classes, having calculated this angle \(2 \pi /3\) or 120 degrees, we can calculate: \(120/11.25 \approx 10.6667.\) This tells us that the vector \(\bigl\langle -1, \sqrt{3} \bigr\rangle\) points in the direction between NNW - North-Northwest and NWbN - Northwest-by-North, a little closer to Northwest-by-North.

  Click here to get a pdf file.
  Mouse-over the image for a simplified version.

  

• Here are a compass rose and the unit circle next to each other:

  

  

• I made some progress on calculating volume of an antiprism whose base is a regular \(n\)-gon and with a given height. For that I use the fact that such an antiprism can be split in \(2n\) congruent pieces. Then I just calculate the volume of one piece. It is like advising a birthday party how to cat a cake in the shape of an antiprism.

  Hover your mouse over an image to see the animation.

  

  

  

  

  
  
  

  

  

  

  

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Wednesday, September 24, 2025

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• The information sheet (updated)
• The main sources for today class are:
  • We first study Vectors. I started writing a webpage to present my vision of the topic Vectors in Two and Three Dimensions.
  • In OpenStax Calculus Volume 3 read Section 2.1 Vectors in the Plane and do as many problems as you can from 1 to 44. (This is the primary source for this topic.)
  • In Calculus Textbook read Vectors and Dot Product. You can skip Example 3 from medicine.
• Since we will be working in three-dimensional space, let's begin by reviewing some foundational geometric terminology. The table below pairs common 2D shapes with their 3D counterparts.
  
  

  2-dimensional object:	square	rectangle	parallelogram	circle	disk	ellipse
  3-dimensional object:	cube	cuboid	parallelepiped	sphere	ball	ellipsoid

  Most of these terms should be familiar. I would like to point out the less frequently used term the cuboid, which is the 3D version of a rectangle. You may know it as a "rectangular prism" or a box 📦. I like the term cuboid; the reason: so prominent object should have its own name, not share it with the family of prisms or a packaging item. The cuboid is a convenient, concise and precise name for a very common shape. As the picture below shows, the term cuboid is used in geometry instruction, even at the primary school level.

  
• I recommend that you always check Wikipedia and other web pages related to the topics that we study. Usually Wikipedia pages are very informative. However, the Wikipedia page for cuboid is strange. The definition given for cuboid on this page is far too technical and difficult to understand. This Wikipedia entry lists the following alternative names for this concept: rectangular box, rectangular hexahedron, right rectangular prism, and rectangular parallelepiped.

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