{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.15 * Show that the definition (1.9) of the cross product is equivalent to the elementary definition that\n",
    "r x s is perpendicular to both r and s, with magnitude $ rs \\sin \\theta $ and direction given by the right-hand\n",
    "rule. \\[Hint: It is a fact (though quite hard to prove) that the definition (1.9) is independent of your choice\n",
    "of axes. Therefore you can choose axes so that r points along the x axis and s lies in the xy plane.\\]\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can map any two 3D vectors U and V into the x-y plane by a suitable rotation of the vectors around the origin.  We can therefore align U along the x axis and V into the x-y plane.\n",
    "\n",
    "$$ \\left|\\begin{array}{cc} \\hat{\\mathbf{i}} &  \\hat{\\mathbf{j}}&  \\hat{\\mathbf{k}}\\\\U_x&0&0\\\\V_x&V_y&0\\end{array}\\right| = U_x V_y \\hat{\\mathbf{k}}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$ \\sin \\theta = \\frac{V_y}{\\sqrt{V_x^2+V_y^2}} $ and therefore $ \\left | U \\right | \\left | V \\right | \\sin \\theta = U_x \\sqrt{V_x^2 + V_y^2} \\frac{V_y}{\\sqrt{V_x^2+V_y^2}} = U_x V_y $ which proves that the determinant version of the cross product and the $ \\left | U \\right | \\left | V \\right | \\sin \\theta $ version are equivalent.\n",
    "\n",
    "Regarding direction if $ V_y $ and $ U_x $ are both positive the U is clockwise of V.  If $ V_y $ and $ U_x $ are both negative then the same is true.  For the remaining cases U is anti-clockwise of V which is in keeping with the right hand rule."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.16 ** (a) Defining the scalar product $ \\mathbf{r \\cdot s} $ by Equation (1.7), $ \\mathbf{r} \\cdot \\mathbf{s} = \\sum r_i s_i $ , show that Pythagoras's theorem implies that the magnitude of any vector r is \n",
    "$ r = \\sqrt{\\mathbf{r} \\cdot \\mathbf{r}} $. (b) It is clear that the length of a\n",
    "vector does not depend on our choice of coordinate axes. Thus the result of part (a) guarantees that the\n",
    "scalar product $ \\mathbf{r} \\cdot \\mathbf{r} $, as defined by (1.7), is the same for any choice of orthogonal axes. Use this to prove\n",
    "that $ \\mathbf{r} \\cdot \\mathbf{s} $, as defined by (1.7), is the same for any choice of orthogonal axes. \\[Hint: Consider the length of the vector $ \\mathbf{r} + \\mathbf{s} $ \\]\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.17 ** (a) Prove that the vector product $ \\mathbf{r \\times s} $ as defined by (1.9) is distributive; that is, that \n",
    "$$ \\mathbf{r \\times (u + v)} = (\\mathbf{r} \\times \\mathbf{u}) + (\\mathbf{r} \\times \\mathbf{v})  $$\n",
    "\n",
    "(b) Prove the product rule\n",
    "$$ \\frac{d}{dt} ( \\mathbf{r \\times s}) = \\mathbf{r} \\times \\frac{d \\mathbf{s}}{dt} + \\frac{d \\mathbf{r}}{dt} \\times  \\mathbf{s}   $$\n",
    "\n",
    "Be careful with the order of the factors.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.18 ** The three vectors a, b, c are the three sides of the triangle ABC with angles $ \\alpha, \\beta, \\gamma $ as shown\n",
    "in Figure 1.15. (a) Prove that the area of the triangle is given by any one of these three expressions:\n",
    "$$ area = \\frac{1}{2} \\left | \\mathbf{ a \\times b} \\right |  =  \\frac{1}{2} \\left | \\mathbf{ b \\times c} \\right | =  \\frac{1}{2} \\left | \\mathbf{ c \\times a} \\right |$$\n",
    " \n",
    "(b) Use the equality of these three expressions to prove the so-called law of sines, that\n",
    "\n",
    "$$ \\frac{a}{\\sin \\alpha} = \\frac{b}{\\sin \\beta} = \\frac{c}{\\sin \\gamma} $$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.19 ** If $ \\mathbf{r, v, a} $ denote the position, velocity, and acceleration of a particle, prove that\n",
    "    \n",
    "$$ \\frac{d}{dt} [ \\mathbf{ a \\cdot ( v \\times r ) } ] = \\mathbf {\\dot{a} \\cdot (v \\times r)} $$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1.20 ** The three vectors A, B, C point from the origin 0 to the three corners of a triangle. Use the\n",
    "result of Problem 1.18 to show that the area of the triangle is given by\n",
    "$$ (area \\ of \\ triangle) = \\frac{1}{2} \\left| \\mathbf{(B \\times C) + (C \\times A) + (A \\times B)} \\right  | $$\n"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "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": "3.6.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}