From 177c6fc121b98df28b0bffe6a5168fe653391870 Mon Sep 17 00:00:00 2001
From: "Kangwon Lee (Education)" <kangwonlee@users.noreply.github.com>
Date: Mon, 13 May 2024 13:07:04 +0900
Subject: [PATCH] 40 / 20 : add b x a

---
 .../20_vector_dot_cross_product.ipynb         | 1660 +++++++++--------
 1 file changed, 920 insertions(+), 740 deletions(-)

diff --git a/40_linear_algebra_1/20_vector_dot_cross_product.ipynb b/40_linear_algebra_1/20_vector_dot_cross_product.ipynb
index d385ef5c..d7e5f78b 100644
--- a/40_linear_algebra_1/20_vector_dot_cross_product.ipynb
+++ b/40_linear_algebra_1/20_vector_dot_cross_product.ipynb
@@ -1,742 +1,922 @@
 {
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "id": "view-in-github",
-    "colab_type": "text"
-   },
-   "source": [
-    "<a href=\"https://colab.research.google.com/github/kangwonlee/nmisp/blob/main/40_linear_algebra_1/20_vector_dot_cross_product.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# 그래프, 수학 기능 추가\n",
-    "# Add graph and math features\n",
-    "import pylab as py\n",
-    "import numpy as np\n",
-    "import numpy.linalg as nl\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# 벡터의 내적과 외적<br>Inner and Cross Products of Vectors\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## 벡터의 내적<br>Inner Product of Vectors\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### `numpy`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "벡터의 내적에 관해 아래 비디오 링크를 참고할 수 있다. (9:09) (내적 주요 설명 1:54 ~ 4:40)<br>\n",
-    "Regarding the vector inner product, you can check the following link. (9:09) (Inner product explanation 1:54 ~ 4:40)<br>\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "[![Vector dot product and vector length 벡터 내적과 벡터의 길이](https://i.ytimg.com/vi/WNuIhXo39_k/hqdefault.jpg)](https://www.youtube.com/watch?v=WNuIhXo39_k)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "예를 들어 2차원 벡터를 생각해 보자.<br>\n",
-    "For example, let's think about two-dimensional vectors (2D vectors).\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "* Feel free to modify the vector definitions below and re-run the cells to explore how different vectors behave under inner and cross product operations.<br>\n",
-    "아래 정의한 벡터는 가능한 예 가운데 하나이므로, 자신의 벡터를 입력하고 해당 노트북을 재실행하여 어떻게 달라지는지 관찰해 보기 바람\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# ref : https://www.youtube.com/watch?v=8QihetGj3pg\n",
-    "b = np.array((2, 5))\n",
-    "c = np.array((7, 1))\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "b\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "c\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "위 2차원 벡터를 한번 그려 보자<br>\n",
-    "Let's plot the 2D vectors above.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def draw_2dvec(x, y, x0=0, y0=0, color='k', name=None):\n",
-    "    py.quiver(x0, y0, x, y, color=color, angles='xy', scale_units='xy', scale=1)\n",
-    "    py.plot((x0, x0+x), (y0, y0+y), alpha=0)\n",
-    "    if name is not None:\n",
-    "        if not name.startswith('$'):\n",
-    "            vec_str = '$\\\\vec{%s}$' % name\n",
-    "        else:\n",
-    "            vec_str = name\n",
-    "        py.text(0.5 * x + x0, 0.5 * y + y0, vec_str)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "scrolled": true
-   },
-   "outputs": [],
-   "source": [
-    "draw_2dvec(b[0], b[1], name='b')\n",
-    "draw_2dvec(c[0], c[1], name='c')\n",
-    "\n",
-    "py.axis('equal')\n",
-    "py.grid(True)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "이 두 벡터의 내적을 계산해 보자<br>Let's calculate the inner product of these two vectors.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "scrolled": true
-   },
-   "outputs": [],
-   "source": [
-    "py.dot(b, c)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "확인해 보자.<br>Let's verify.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "b[0]*c[0] + b[1]*c[1]\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "다른 방법도 있다.<br>Other methods are available.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# https://docs.scipy.org/doc/numpy/reference/generated/numpy.inner.html\n",
-    "py.inner(b, c)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# https://docs.scipy.org/doc/numpy/reference/generated/numpy.inner.html\n",
-    "sum(b[:] * c[:])\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "**Python 3.5, Numpy 1.10 이상**에서는 **`@` 연산자**를 내적의 의미로 사용할 수 있다.<br>\n",
-    "For **Python 3.5 or higher and Numpy 1.10 or higher**, we can use **`@` operator** as inner product.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "b @ c\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "theta_a_deg = 30\n",
-    "theta_a_rad = np.deg2rad(theta_a_deg)\n",
-    "theta_b_deg = 120\n",
-    "theta_b_rad = np.deg2rad(theta_b_deg)\n",
-    "\n",
-    "a_row_array = np.array([np.cos(theta_a_rad), np.sin(theta_a_rad)])\n",
-    "b_row_array = np.array([np.cos(theta_b_rad), np.sin(theta_b_rad)])\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "draw_2dvec(a_row_array[0], a_row_array[1], name='a_{array}')\n",
-    "draw_2dvec(b_row_array[0], b_row_array[1], name='b_{array}')\n",
-    "\n",
-    "py.axis('equal')\n",
-    "py.grid(True)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "print('a dot b (array) =', np.dot(a_row_array, b_row_array))\n",
-    "print('a_row_array @ b_row_array =', a_row_array @ b_row_array)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "* What happens to the inner product if you change the angle between the vectors?<br>벡터 사잇각이 달라지면 내적에는 어떤 영향을 미치는가?\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### 표준 기능<br>Standard library\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "b_list = [2, 5]\n",
-    "c_list = [7, 1]\n",
-    "print(f\"b_list = {b_list}\")\n",
-    "print(f\"c_list = {c_list}\")\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def dot_product_list_comprehension(x_list, y_list):\n",
-    "    assert len(x_list) == len(y_list)\n",
-    "    return sum(xi*yi for xi, yi in zip(x_list, y_list))\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def dot_product_map(x_list, y_list):\n",
-    "    assert len(x_list) == len(y_list)\n",
-    "    return sum(map(lambda x_y:x_y[0] * x_y[1],zip(x_list, y_list)))\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def dot_product_list_for(x_list, y_list):\n",
-    "    assert len(x_list) == len(y_list)\n",
-    "    s = 0\n",
-    "    for xi, yi in zip(x_list, y_list):\n",
-    "        s += (xi * yi)\n",
-    "    return s\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def dot_product_list_for_index(x_list, y_list):\n",
-    "    assert len(x_list) == len(y_list)\n",
-    "    s = 0\n",
-    "    n = len(x_list)\n",
-    "    for i in range(n):\n",
-    "        s += (x_list[i] * y_list[i])\n",
-    "    return s\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "print(f\"dot_product_list_comprehension(b_list, c_list) = {dot_product_list_comprehension(b_list, c_list)}\")\n",
-    "print(f\"dot_product_map(b_list, c_list) = {dot_product_map(b_list, c_list)}\")\n",
-    "print(f\"dot_product_list_for(b_list, c_list) = {dot_product_list_for(b_list, c_list)}\")\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "assert (\n",
-    "    dot_product_list_comprehension(b_list, c_list) \n",
-    "    == dot_product_map(b_list, c_list)\n",
-    "    == dot_product_list_for(b_list, c_list)\n",
-    ")\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## 벡터의 외적<br>Cross Product of Vectors\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "아래 비디오 링크를 참고할 수 있다. (15:46)<br>\n",
-    "Regarding the vector inner product, you can check the following link. (15:46)<br>\n",
-    "<br>\n",
-    "[![칸 아카데미 | 벡터와 벡터 공간 | 외적 소개<br>Khan Academy | Linear Algebra | Vectors and spaces | Cross product introduction](https://i.ytimg.com/vi/pJzmiywagfY/hqdefault.jpg)](https://www.youtube.com/watch?v=pJzmiywagfY)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "열벡터의 잇점은 무엇이라고 생각하는가?<br>\n",
-    "What do you think is the advantage of the column vector form?\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### `numpy`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "다음과 같은 두 벡터를 생각해 보자.<br>Let's think about following two vectors.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "a = py.array((1, -7, 1))\n",
-    "b = py.array((5, 2, 4))\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "외적을 계산해 보자.<br>Let's calculate the cross product.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "a_cross_b = py.cross(a, b)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "a_cross_b\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "확인 해 보자.<br>Let's verify.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "py.dot(a, a_cross_b)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "py.dot(b, a_cross_b)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "외적의 결과가 $\\vec{a}$, $\\vec{b}$ 와 수직임을 알 수 있다.<br>We can see that the cross product is orthogonal to $\\vec{a}$ and $\\vec{b}$.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "from mpl_toolkits.mplot3d import Axes3D\n",
-    "import numpy as np\n",
-    "\n",
-    "\n",
-    "def plot_3dvec(ax, xyz, name=None):\n",
-    "    ax.quiver(\n",
-    "      *[0, 0, 0], *xyz,\n",
-    "    )\n",
-    "\n",
-    "    ax.plot3D(\n",
-    "        (0, xyz[0]),\n",
-    "        (0, xyz[1]),\n",
-    "        (0, xyz[2]),\n",
-    "        alpha=0\n",
-    "    )\n",
-    "\n",
-    "    if name is not None:\n",
-    "        ax.text3D(*xyz, name)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "fig = plt.figure()\n",
-    "ax = fig.add_subplot(projection='3d')\n",
-    "\n",
-    "plot_3dvec(ax, a, r'$\\vec{a}$')\n",
-    "plot_3dvec(ax, b, r'$\\vec{b}$')\n",
-    "plot_3dvec(ax, a_cross_b, r'$\\vec{cross}$')\n",
-    "\n",
-    "ax.grid(True)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "* How does the cross product change if you reverse the order of the vectors?<br>외적의 순서를 바꾸면 결과에 어떤 영향이 있는가?\n",
-    "* Can you find two vectors whose cross product is the zero vector? What does this mean geometrically?<br>외적을 영벡터로 만드는 두 벡터를 찾을 수 있는가? 기하학적으로는 어떤 의미인가?\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### 표준기능<br>Standard library\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "a_list = (1, -7, 1)\n",
-    "b_list = (5, 2, 4)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def cross_product_list(x_list, y_list):\n",
-    "    assert 3 == len(x_list)\n",
-    "    assert 3 == len(y_list)\n",
-    "    return [\n",
-    "        x_list[1]*y_list[2] - x_list[2]*y_list[1],\n",
-    "        x_list[0]*y_list[2] - x_list[2]*y_list[0],\n",
-    "        x_list[0]*y_list[1] - x_list[1]*y_list[0],\n",
-    "    ]\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "a_cross_b_stdlib = cross_product_list(a_list, b_list)\n",
-    "print(f\"cross_product_list(a_list, b_list) = {a_cross_b_stdlib}\")\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "fig = plt.figure()\n",
-    "ax = fig.add_subplot(projection='3d')\n",
-    "\n",
-    "plot_3dvec(ax, a_list, r'$\\vec{a}$')\n",
-    "plot_3dvec(ax, b_list, r'$\\vec{b}$')\n",
-    "plot_3dvec(ax, a_cross_b_stdlib, r'$\\vec{cross}$')\n",
-    "\n",
-    "ax.grid(True)\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## 연습 문제<br>Exercise\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "Try This 1: Given a force vector $\\vec{f}=(10, 10)$, find its components along the directions of the vectors $\\vec{x'}, \\vec{y'}$ directions.<br>힘 벡터 (10, 10) 의 $\\vec{x'}$ and $\\vec{y'}$ 방향 성분을 각각 구하시오.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "$$\n",
-    "\\begin{align}\n",
-    "    \\vec{x'}&=\\left(0.8, 0.6\\right) \\\\\n",
-    "    \\vec{y'}&=\\left(-0.6, 0.8\\right)\n",
-    "\\end{align}\n",
-    "$$\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "Try This 2: Find the magnitude and direction of the moment produced by a force vector  \n",
-    "$\\vec{f}=(-1, 1, -1)$ acting at the point $\\vec{r}= (1, 1, 1)$ on the origin.<br>힘의 작용점 $\\vec{r}= (1, 1, 1)$인 힘 벡터 $\\vec{f}=(-1, 1, -1)$가 원점에 작용하는 모멘트의 크기와 방향을 구하시오\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Final Bell<br>마지막 종\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# stackoverfow.com/a/24634221\n",
-    "import os\n",
-    "os.system(\"printf '\\a'\");\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "colab": {
-   "include_colab_link": true,
-   "provenance": []
-  },
-  "kernelspec": {
-   "display_name": "Python 3 (ipykernel)",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.0"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 4
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "view-in-github",
+        "colab_type": "text"
+      },
+      "source": [
+        "<a href=\"https://colab.research.google.com/github/kangwonlee/nmisp/blob/main/40_linear_algebra_1/20_vector_dot_cross_product.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "Aj2lSYkS500L"
+      },
+      "outputs": [],
+      "source": [
+        "# 그래프, 수학 기능 추가\n",
+        "# Add graph and math features\n",
+        "import pylab as py\n",
+        "import numpy as np\n",
+        "import numpy.linalg as nl\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "iF6em7B9500P"
+      },
+      "source": [
+        "# 벡터의 내적과 외적<br>Inner and Cross Products of Vectors\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "gmd27txO500R"
+      },
+      "source": [
+        "## 벡터의 내적<br>Inner Product of Vectors\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "VhkkOlyO500R"
+      },
+      "source": [
+        "### `numpy`\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "F4AF7Xxl500S"
+      },
+      "source": [
+        "벡터의 내적에 관해 아래 비디오 링크를 참고할 수 있다. (9:09) (내적 주요 설명 1:54 ~ 4:40)<br>\n",
+        "Regarding the vector inner product, you can check the following link. (9:09) (Inner product explanation 1:54 ~ 4:40)<br>\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "vMMKHNWe500S"
+      },
+      "source": [
+        "[![Vector dot product and vector length 벡터 내적과 벡터의 길이](https://i.ytimg.com/vi/WNuIhXo39_k/hqdefault.jpg)](https://www.youtube.com/watch?v=WNuIhXo39_k)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "gTvSAAfF500T"
+      },
+      "source": [
+        "예를 들어 2차원 벡터를 생각해 보자.<br>\n",
+        "For example, let's think about two-dimensional vectors (2D vectors).\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "VwzvVaZx500T"
+      },
+      "source": [
+        "* Feel free to modify the vector definitions below and re-run the cells to explore how different vectors behave under inner and cross product operations.<br>\n",
+        "아래 정의한 벡터는 가능한 예 가운데 하나이므로, 자신의 벡터를 입력하고 해당 노트북을 재실행하여 어떻게 달라지는지 관찰해 보기 바람\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "p3QTTZFe500U"
+      },
+      "outputs": [],
+      "source": [
+        "# ref : https://www.youtube.com/watch?v=8QihetGj3pg\n",
+        "b = np.array((2, 5))\n",
+        "c = np.array((7, 1))\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "LWtftjpq500U"
+      },
+      "outputs": [],
+      "source": [
+        "b\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "O63Ziuph500U"
+      },
+      "outputs": [],
+      "source": [
+        "c\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "yLhNGDdh500V"
+      },
+      "source": [
+        "위 2차원 벡터를 한번 그려 보자<br>\n",
+        "Let's plot the 2D vectors above.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "NBDsiTnl500V"
+      },
+      "outputs": [],
+      "source": [
+        "def draw_2dvec(x, y, x0=0, y0=0, color='k', name=None):\n",
+        "    py.quiver(x0, y0, x, y, color=color, angles='xy', scale_units='xy', scale=1)\n",
+        "    py.plot((x0, x0+x), (y0, y0+y), alpha=0)\n",
+        "    if name is not None:\n",
+        "        if not name.startswith('$'):\n",
+        "            vec_str = '$\\\\vec{%s}$' % name\n",
+        "        else:\n",
+        "            vec_str = name\n",
+        "        py.text(0.5 * x + x0, 0.5 * y + y0, vec_str)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "scrolled": true,
+        "id": "1cINeIrW500V"
+      },
+      "outputs": [],
+      "source": [
+        "draw_2dvec(b[0], b[1], name='b')\n",
+        "draw_2dvec(c[0], c[1], name='c')\n",
+        "\n",
+        "py.axis('equal')\n",
+        "py.grid(True)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "9tVwWnAy500V"
+      },
+      "source": [
+        "이 두 벡터의 내적을 계산해 보자<br>Let's calculate the inner product of these two vectors.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "scrolled": true,
+        "id": "i42hzJ2t500W"
+      },
+      "outputs": [],
+      "source": [
+        "py.dot(b, c)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "d66QgCaa500W"
+      },
+      "source": [
+        "확인해 보자.<br>Let's verify.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "UMCB8LXu500W"
+      },
+      "outputs": [],
+      "source": [
+        "b[0]*c[0] + b[1]*c[1]\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "_nLgBB89500W"
+      },
+      "source": [
+        "다른 방법도 있다.<br>Other methods are available.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "4khKjpcn500W"
+      },
+      "outputs": [],
+      "source": [
+        "# https://docs.scipy.org/doc/numpy/reference/generated/numpy.inner.html\n",
+        "py.inner(b, c)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "gO_2eFpP500W"
+      },
+      "outputs": [],
+      "source": [
+        "# https://docs.scipy.org/doc/numpy/reference/generated/numpy.inner.html\n",
+        "sum(b[:] * c[:])\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "eBhAZH5F500W"
+      },
+      "source": [
+        "**Python 3.5, Numpy 1.10 이상**에서는 **`@` 연산자**를 내적의 의미로 사용할 수 있다.<br>\n",
+        "For **Python 3.5 or higher and Numpy 1.10 or higher**, we can use **`@` operator** as inner product.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "2Gl2eKJX500W"
+      },
+      "outputs": [],
+      "source": [
+        "b @ c\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "rCssHIta500X"
+      },
+      "outputs": [],
+      "source": [
+        "theta_a_deg = 30\n",
+        "theta_a_rad = np.deg2rad(theta_a_deg)\n",
+        "theta_b_deg = 120\n",
+        "theta_b_rad = np.deg2rad(theta_b_deg)\n",
+        "\n",
+        "a_row_array = np.array([np.cos(theta_a_rad), np.sin(theta_a_rad)])\n",
+        "b_row_array = np.array([np.cos(theta_b_rad), np.sin(theta_b_rad)])\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "VkLyv6tC500X"
+      },
+      "outputs": [],
+      "source": [
+        "draw_2dvec(a_row_array[0], a_row_array[1], name='a_{array}')\n",
+        "draw_2dvec(b_row_array[0], b_row_array[1], name='b_{array}')\n",
+        "\n",
+        "py.axis('equal')\n",
+        "py.grid(True)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "VfI3mxAm500X"
+      },
+      "outputs": [],
+      "source": [
+        "print('a dot b (array) =', np.dot(a_row_array, b_row_array))\n",
+        "print('a_row_array @ b_row_array =', a_row_array @ b_row_array)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "dOdn8BZa500X"
+      },
+      "source": [
+        "* What happens to the inner product if you change the angle between the vectors?<br>벡터 사잇각이 달라지면 내적에는 어떤 영향을 미치는가?\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "8JxJYmNy500X"
+      },
+      "source": [
+        "### 표준 기능<br>Standard library\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "Rmlj6P1o500X"
+      },
+      "outputs": [],
+      "source": [
+        "b_list = [2, 5]\n",
+        "c_list = [7, 1]\n",
+        "print(f\"b_list = {b_list}\")\n",
+        "print(f\"c_list = {c_list}\")\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "Y2eQBR2U500X"
+      },
+      "outputs": [],
+      "source": [
+        "def dot_product_list_comprehension(x_list, y_list):\n",
+        "    assert len(x_list) == len(y_list)\n",
+        "    return sum(xi*yi for xi, yi in zip(x_list, y_list))\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "YDt4f84E500X"
+      },
+      "outputs": [],
+      "source": [
+        "def dot_product_map(x_list, y_list):\n",
+        "    assert len(x_list) == len(y_list)\n",
+        "    return sum(map(lambda x_y:x_y[0] * x_y[1],zip(x_list, y_list)))\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "_wJXNMGD500X"
+      },
+      "outputs": [],
+      "source": [
+        "def dot_product_list_for(x_list, y_list):\n",
+        "    assert len(x_list) == len(y_list)\n",
+        "    s = 0\n",
+        "    for xi, yi in zip(x_list, y_list):\n",
+        "        s += (xi * yi)\n",
+        "    return s\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "0EGXAiSj500Y"
+      },
+      "outputs": [],
+      "source": [
+        "def dot_product_list_for_index(x_list, y_list):\n",
+        "    assert len(x_list) == len(y_list)\n",
+        "    s = 0\n",
+        "    n = len(x_list)\n",
+        "    for i in range(n):\n",
+        "        s += (x_list[i] * y_list[i])\n",
+        "    return s\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "F4vE22oz500Y"
+      },
+      "outputs": [],
+      "source": [
+        "print(f\"dot_product_list_comprehension(b_list, c_list) = {dot_product_list_comprehension(b_list, c_list)}\")\n",
+        "print(f\"dot_product_map(b_list, c_list) = {dot_product_map(b_list, c_list)}\")\n",
+        "print(f\"dot_product_list_for(b_list, c_list) = {dot_product_list_for(b_list, c_list)}\")\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "uXpnr9x4500Y"
+      },
+      "outputs": [],
+      "source": [
+        "assert (\n",
+        "    dot_product_list_comprehension(b_list, c_list)\n",
+        "    == dot_product_map(b_list, c_list)\n",
+        "    == dot_product_list_for(b_list, c_list)\n",
+        ")\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "rdelZ6MH500Y"
+      },
+      "source": [
+        "## 벡터의 외적<br>Cross Product of Vectors\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "-HPI5L8P500Y"
+      },
+      "source": [
+        "아래 비디오 링크를 참고할 수 있다. (15:46)<br>\n",
+        "Regarding the vector inner product, you can check the following link. (15:46)<br>\n",
+        "<br>\n",
+        "[![칸 아카데미 | 벡터와 벡터 공간 | 외적 소개<br>Khan Academy | Linear Algebra | Vectors and spaces | Cross product introduction](https://i.ytimg.com/vi/pJzmiywagfY/hqdefault.jpg)](https://www.youtube.com/watch?v=pJzmiywagfY)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "8iflc6pX500Y"
+      },
+      "source": [
+        "열벡터의 잇점은 무엇이라고 생각하는가?<br>\n",
+        "What do you think is the advantage of the column vector form?\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "bqNG6DZK500Y"
+      },
+      "source": [
+        "### `numpy`\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "kIFDvMPJ500c"
+      },
+      "source": [
+        "다음과 같은 두 벡터를 생각해 보자.<br>Let's think about following two vectors.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "WtpfsI7w500c"
+      },
+      "outputs": [],
+      "source": [
+        "a = py.array((1, -7, 1))\n",
+        "b = py.array((5, 2, 4))\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "h2sr9nRp500c"
+      },
+      "source": [
+        "외적을 계산해 보자.<br>Let's calculate the cross product.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "3Tj5yMNi500c"
+      },
+      "outputs": [],
+      "source": [
+        "a_cross_b = py.cross(a, b)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "DaE8z01N500c"
+      },
+      "outputs": [],
+      "source": [
+        "a_cross_b\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "RIicfHe0500c"
+      },
+      "source": [
+        "확인 해 보자.<br>Let's verify.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "BCJBM_k7500c"
+      },
+      "outputs": [],
+      "source": [
+        "py.dot(a, a_cross_b)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "_A6XgNLF500d"
+      },
+      "outputs": [],
+      "source": [
+        "py.dot(b, a_cross_b)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "7euVhGhZ500d"
+      },
+      "source": [
+        "외적의 결과가 $\\vec{a}$, $\\vec{b}$ 와 수직임을 알 수 있다.<br>We can see that the cross product is orthogonal to $\\vec{a}$ and $\\vec{b}$.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "2F2QAtEP500d"
+      },
+      "outputs": [],
+      "source": [
+        "import matplotlib.pyplot as plt\n",
+        "from mpl_toolkits.mplot3d import Axes3D\n",
+        "import numpy as np\n",
+        "\n",
+        "\n",
+        "def plot_3dvec(ax, xyz, name=None):\n",
+        "    ax.quiver(\n",
+        "      *[0, 0, 0], *xyz,\n",
+        "    )\n",
+        "\n",
+        "    ax.plot3D(\n",
+        "        (0, xyz[0]),\n",
+        "        (0, xyz[1]),\n",
+        "        (0, xyz[2]),\n",
+        "        alpha=0\n",
+        "    )\n",
+        "\n",
+        "    if name is not None:\n",
+        "        ax.text3D(*xyz, name)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "yPP4Mm-f500d"
+      },
+      "outputs": [],
+      "source": [
+        "fig = plt.figure()\n",
+        "ax = fig.add_subplot(projection='3d')\n",
+        "\n",
+        "plot_3dvec(ax, a, r'$\\vec{a}$')\n",
+        "plot_3dvec(ax, b, r'$\\vec{b}$')\n",
+        "plot_3dvec(ax, a_cross_b, r'$\\vec{cross}$')\n",
+        "\n",
+        "ax.grid(True)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "lrD0rUAs500d"
+      },
+      "source": [
+        "* How does the cross product change if you reverse the order of the vectors?<br>외적의 순서를 바꾸면 결과에 어떤 영향이 있는가?\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "IrSBJ7wIHtiH"
+      },
+      "outputs": [],
+      "source": [
+        "b_cross_a = py.cross(b, a)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "C9_ITNXHHtiQ"
+      },
+      "outputs": [],
+      "source": [
+        "b_cross_a\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "source": [
+        "fig = plt.figure()\n",
+        "ax = fig.add_subplot(projection='3d')\n",
+        "\n",
+        "plot_3dvec(ax, b, r'$\\vec{b}$')\n",
+        "plot_3dvec(ax, a, r'$\\vec{a}$')\n",
+        "plot_3dvec(ax, b_cross_a, r'$\\vec{cross}$')\n",
+        "\n",
+        "ax.grid(True)\n",
+        "\n"
+      ],
+      "metadata": {
+        "id": "17yrlGc5H77Y"
+      },
+      "execution_count": null,
+      "outputs": []
+    },
+    {
+      "cell_type": "markdown",
+      "source": [
+        "* Can you find two vectors whose cross product is the zero vector? What does this mean geometrically?<br>외적을 영벡터로 만드는 두 벡터를 찾을 수 있는가? 기하학적으로는 어떤 의미인가?\n",
+        "\n"
+      ],
+      "metadata": {
+        "id": "I1C5LwHSHyoQ"
+      }
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "8ujwIG7i500d"
+      },
+      "source": [
+        "### 표준기능<br>Standard library\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "wgUeJyAI500d"
+      },
+      "outputs": [],
+      "source": [
+        "a_list = (1, -7, 1)\n",
+        "b_list = (5, 2, 4)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "g6wCS8K_500d"
+      },
+      "outputs": [],
+      "source": [
+        "def cross_product_list(x_list, y_list):\n",
+        "    assert 3 == len(x_list)\n",
+        "    assert 3 == len(y_list)\n",
+        "    return [\n",
+        "        x_list[1]*y_list[2] - x_list[2]*y_list[1],\n",
+        "        x_list[0]*y_list[2] - x_list[2]*y_list[0],\n",
+        "        x_list[0]*y_list[1] - x_list[1]*y_list[0],\n",
+        "    ]\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "8f8sEu9m500e"
+      },
+      "outputs": [],
+      "source": [
+        "a_cross_b_stdlib = cross_product_list(a_list, b_list)\n",
+        "print(f\"cross_product_list(a_list, b_list) = {a_cross_b_stdlib}\")\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "PJj3hPIT500e"
+      },
+      "outputs": [],
+      "source": [
+        "fig = plt.figure()\n",
+        "ax = fig.add_subplot(projection='3d')\n",
+        "\n",
+        "plot_3dvec(ax, a_list, r'$\\vec{a}$')\n",
+        "plot_3dvec(ax, b_list, r'$\\vec{b}$')\n",
+        "plot_3dvec(ax, a_cross_b_stdlib, r'$\\vec{cross}$')\n",
+        "\n",
+        "ax.grid(True)\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "eYNSqPbf500e"
+      },
+      "source": [
+        "## 연습 문제<br>Exercise\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "6hqqWpze500e"
+      },
+      "source": [
+        "Try This 1: Given a force vector $\\vec{f}=(10, 10)$, find its components along the directions of the vectors $\\vec{x'}, \\vec{y'}$ directions.<br>힘 벡터 (10, 10) 의 $\\vec{x'}$ and $\\vec{y'}$ 방향 성분을 각각 구하시오.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "GS1AxWDs500e"
+      },
+      "source": [
+        "$$\n",
+        "\\begin{align}\n",
+        "    \\vec{x'}&=\\left(0.8, 0.6\\right) \\\\\n",
+        "    \\vec{y'}&=\\left(-0.6, 0.8\\right)\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "caYsn2eA500e"
+      },
+      "outputs": [],
+      "source": []
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "6YiosVCf500e"
+      },
+      "source": [
+        "Try This 2: Find the magnitude and direction of the moment produced by a force vector  \n",
+        "$\\vec{f}=(-1, 1, -1)$ acting at the point $\\vec{r}= (1, 1, 1)$ on the origin.<br>힘의 작용점 $\\vec{r}= (1, 1, 1)$인 힘 벡터 $\\vec{f}=(-1, 1, -1)$가 원점에 작용하는 모멘트의 크기와 방향을 구하시오\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "DyA5LSiu500e"
+      },
+      "outputs": [],
+      "source": []
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "29XfZE4Z500f"
+      },
+      "source": [
+        "## Final Bell<br>마지막 종\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "VUqpsw-k500g"
+      },
+      "outputs": [],
+      "source": [
+        "# stackoverfow.com/a/24634221\n",
+        "import os\n",
+        "os.system(\"printf '\\a'\");\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "metadata": {
+        "id": "xtD6gRLx500g"
+      },
+      "outputs": [],
+      "source": []
+    }
+  ],
+  "metadata": {
+    "colab": {
+      "provenance": [],
+      "include_colab_link": true
+    },
+    "kernelspec": {
+      "display_name": "Python 3 (ipykernel)",
+      "language": "python",
+      "name": "python3"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "ipython",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.10.0"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 0
 }
\ No newline at end of file