Math_Vector_Basic01.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 1: 다음에 언급된 Physical Quantity들이 scalar인지 vector인지 고르시오.\n",
    "\n",
    "1. $20m^2$의 넓이 : Scalar\n",
    "2. $10N$의 힘(F) : Vector"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 2: 다음 벡터의 2-Norm(크기)을 구하시오\n",
    "\n",
    "1. $ \\vec{a} = <1,0,2>$\n",
    "2. $ \\vec{b}=<0,3>$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2-1: 2.23606797749979 |np.sqrt(5) 2.23606797749979\n",
      "2-2: 3.0\n"
     ]
    }
   ],
   "source": [
    "vec = [[1,0,2],[0,3,0]]\n",
    "L2_norm = np.linalg.norm(vec,axis=1,ord=2)\n",
    "print('2-1:',L2_norm[0],'|np.sqrt(5)',np.sqrt(5))\n",
    "print('2-2:',L2_norm[1])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 3: 다음을 계산하시오.\n",
    "\n",
    "다음과 같은 vector가 있음.\n",
    "\n",
    "$$\n",
    "\\vec{a}=<2,1,-3> \\\\\n",
    "\\vec{b}=<0,4,-2>\n",
    "$$\n",
    "\n",
    "1. $\\vec{a}+\\vec{b}$\n",
    "2. $\\vec{a}-2\\vec{b}$\n",
    "3. $\\vec{a}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "3-1: [ 2  5 -5]\n",
      "3-2: [ 2 -7  1]\n",
      "3-3: [ 2  1 -3]\n"
     ]
    }
   ],
   "source": [
    "a = np.array([2,1,-3])\n",
    "b = np.array([0,4,-2])\n",
    "\n",
    "print('3-1:',a+b)\n",
    "print('3-2:',a-2*b)\n",
    "print('3-3:',a)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 4: 다음 중 Unit vector인 것은?\n",
    "\n",
    "1. $\\vec{a}=<0,2,-1>$\n",
    "2. $\\vec{a}=<0,0,-1>$\n",
    "3. $\\vec{a}=<1,1,1>$\n",
    "4. $\\vec{a}=<\\frac{3}{5},0,\\frac{4}{5}>$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "4-1:False\n",
      "4-2:True\n",
      "4-3:False\n",
      "4-4:True\n"
     ]
    }
   ],
   "source": [
    "vec = [[0,2,-1],[0,0,-1],[1,1,1],[3/5,0,4/5]]\n",
    "L2_norm = np.linalg.norm(vec,axis=1,ord=2)\n",
    "answers = [L2_norm==1]\n",
    "for idx,val in enumerate(answers[0]):\n",
    "    print('4-{}:{}'.format(idx+1,val))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 5: 다음의 vector와 반대 방향의 unit vector를 구하시오.\n",
    "\n",
    "$$ \\vec{a}=<3,1,-2> $$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[-0.80178373 -0.26726124  0.53452248]\n",
      "-0.8017837257372732 -0.2672612419124244 0.5345224838248488\n"
     ]
    }
   ],
   "source": [
    "a = np.array([3,1,-2])\n",
    "b = -1*a/np.linalg.norm(a)\n",
    "print(b)\n",
    "print(-3/np.sqrt(14),-1/np.sqrt(14),2/np.sqrt(14))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 6: 다음을 구하시오\n",
    "\n",
    "$$ \\vec{a}=<3,-2,1> \\\\ \\vec{b}=<-4,5,1> $$\n",
    "\n",
    "1. $\\vec{a}\\cdot\\vec{b}$\n",
    "2. $\\vec{b}\\cdot\\vec{b}$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6-1: -21\n",
      "6-2: 42\n"
     ]
    }
   ],
   "source": [
    "a = np.array([3,-2,1])\n",
    "b = np.array([-4,5,1])\n",
    "\n",
    "print('6-1:',np.dot(a,b))\n",
    "print('6-2:',np.dot(b,b))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 7: 다음의 vector들이 orthogonal한지 증명하시오.\n",
    "\n",
    "1. $\\vec{a}=<4,0,0>, \\vec{b}=<0,-3,1>$\n",
    "2. $\\vec{a}=<4,0,1>, \\vec{b}=<0,-2,1>$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "7-1: a and b is orthogonal: True\n",
      "7-2: a and b is orthogonal: False\n"
     ]
    }
   ],
   "source": [
    "def is_orthogonal(a,b):\n",
    "    dot = np.dot(a,b)\n",
    "    if dot == 0:\n",
    "        #print('orthogonal')\n",
    "        return True\n",
    "    else:\n",
    "        #print('not orthogonal')\n",
    "        return False\n",
    "        \n",
    "    \n",
    "a = np.array([4,0,0])\n",
    "b = np.array([0,-3,1])\n",
    "print('7-1: a and b is orthogonal:',is_orthogonal(a,b))\n",
    "\n",
    "a = np.array([4,0,1])\n",
    "b = np.array([0,-2,1])\n",
    "print('7-2: a and b is orthogonal:',is_orthogonal(a,b))\n",
    "\n",
    "#a = np.array([0,0,1])\n",
    "#b = np.array([0,1,0])\n",
    "#print('7-3: a and b is orthogonal:',is_orthogonal(a,b))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 8: 다음을 구하시오.\n",
    "\n",
    "$$ \\vec{a}=<1,2,3> \\\\ \\vec{b}=<4,0,1> $$\n",
    "\n",
    "1. $\\vec{a}\\times\\vec{b}$\n",
    "2. $\\vec{b}\\times\\vec{b}$\n",
    "3. $\\vec{b}\\times\\vec{a}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "8-1: [ 2 11 -8]\n",
      "8-2: [0 0 0]\n",
      "8-3: [ -2 -11   8]\n"
     ]
    }
   ],
   "source": [
    "a = np.array([1,2,3])\n",
    "b = np.array([4,0,1])\n",
    "\n",
    "print('8-1:',np.cross(a,b))\n",
    "print('8-2:',np.cross(b,b))\n",
    "print('8-3:',np.cross(b,a))"
   ]
  }
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