Ch2_Euclidean_Space_Ex_2.2.ipynb

{
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   "source": [
    "<b> Question 1</b> <br>\n",
    "Let $u$ and $v$ be vectors in $\\mathbb{R}^2$. For the following $u$ and $v$ determine the angle between the vectors and label this angle and the vectors.\n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "(b)$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\qquad $\n",
    "(c)$u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$, $v = \\begin{pmatrix}\n",
    "1/2\\\\\n",
    "-1/2\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0ab8035c",
   "metadata": {},
   "source": [
    "<b> Solution </b> <br> \n",
    "<b>(a)</b> <br> $cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$\n",
    "\n",
    "$u \\cdot v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = \\sqrt{1(0) + 1(1)} = 1$ \n",
    "\n",
    "$\\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{1^2 + 2^2 \\cdot} \\sqrt{0^2 + 1^2} = \\sqrt{2}$\n",
    "\n",
    "$ cos(\\theta) = \\frac{1}{\\sqrt2}$<br>\n",
    "$ \\Rightarrow \\theta = 45^o$\n",
    "\n",
    "<b>(b)</b> <br> $cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$ \n",
    "\n",
    "$u \\cdot v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = \\sqrt{1(0) + 0(1)} = 0$\n",
    "\n",
    "Since, $u \\cdot v = 0$ <br>\n",
    "$\\Rightarrow \\theta = 90^o$\n",
    "\n",
    "<b>(c)</b> <br>\n",
    "$cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$ <br>\n",
    "$u \\cdot v = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1/2\\\\\n",
    "-1/2\\\\\n",
    "\\end{pmatrix} = \\sqrt{-2(1/2) + 3(-1/2)} = -5/2$\n",
    "\n",
    "\n",
    "$\\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{(-2)^2 + 3^2} \\cdot \\sqrt{ (1/2)^2 + (-1/2)^2} = \\sqrt{13} \\cdot \\sqrt{\\frac{1}{2}}$\n",
    "\n",
    "$\\Rightarrow cos(\\theta) = \\frac{-5/2}{\\sqrt{13} \\cdot \\sqrt{\\frac{1}{2}}} = -0.9805$ <br>\n",
    "$\\Rightarrow \\theta = cos^{-1}(-0.98)$"
   ]
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    "-------------------------------------------------------------------------------------"
   ]
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   "id": "47275429",
   "metadata": {},
   "source": [
    "<b> Question 2</b> <br>\n",
    "\n",
    "For the following vetors $u$ and $v$ in $\\mathbb{R}^3$ determine the angle between them\n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "3\\\\\n",
    "-1\\\\\n",
    "5\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "(b)$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "15\\\\\n",
    "\\end{pmatrix}\\qquad $\n",
    "(c)$u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$, $v = \\begin{pmatrix}\n",
    "\\sqrt2\\\\\n",
    "1/\\sqrt2\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
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   "id": "382f4515",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "<b>(a)</b><br>\n",
    "$cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$ <br>\n",
    "$u \\cdot v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3\\\\\n",
    "-1\\\\\n",
    "5\\\\\n",
    "\\end{pmatrix} = -1(3) + 1(-1) + 3(5) = -3-1+15 = 11$\n",
    "\n",
    "$\\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{(-1)^2 + 1^2 + 3^2} \\cdot \\sqrt{3^2 + (-1)^2 + 5^2} = \\sqrt{11} \\cdot \\sqrt{35}$\n",
    "\n",
    "$\\Rightarrow cos(\\theta) = \\frac{11}{\\sqrt{11}\\cdot \\sqrt{35}}$\n",
    "\n",
    "\n",
    "$\\Rightarrow \\theta = cos^{-1} 0.5606$\n",
    "\n",
    "<b>(b)</b><br>\n",
    "\n",
    "$ u \\cdot v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "15\\\\\n",
    "\\end{pmatrix}  = 1(0) + 0(0) + 0(15) = 0$ <br>\n",
    "Since, $u \\cdot v = 0$ $u$ and $v$ must be orthogonal<br>\n",
    "$\\Rightarrow \\theta = 90^o$ \n",
    "\n",
    "<b>(c)</b> <br>\n",
    "$u \\cdot v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "\\sqrt2\\\\\n",
    "1/\\sqrt{2}\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} = -1(\\sqrt2) + 2(\\frac{1}{\\sqrt2}) + 3(-1) = -3$ <br>\n",
    "$\\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{(-1)^2 + 2^2 + 3^2} \\cdot \\sqrt{\\sqrt2^2 + (1/\\sqrt{2})^2 + (1)^2} = \\sqrt{14} \\cdot \\sqrt{7/2}$ \n",
    "\n",
    "$cos(\\theta) = \\frac{-3}{\\sqrt{14} \\cdot\\sqrt{\\frac{7}{2}}} = \\frac{-3}{7}$ \n",
    "\n",
    "$\\Rightarrow \\theta = cos^{-1}(\\frac{-3}{7})$"
   ]
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    "------------------------------------------------------------------------------------------------"
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   "cell_type": "markdown",
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   "metadata": {},
   "source": [
    "<b> Question 3</b> <br>\n",
    "\n",
    "Find the angle between the following vectors in $\\mathbb{R}^4$\n",
    "\n",
    "(a)$ u  = \\begin{pmatrix}\n",
    "2&3&-8&1\\\\\n",
    "\\end{pmatrix}^T\\qquad$, $ v  = \\begin{pmatrix}\n",
    "-1&2&-5&-3\\\\\n",
    "\\end{pmatrix}^T$<br>\n",
    "(b)$ u  = \\begin{pmatrix}\n",
    "-2&-3&-1&-1\\\\\n",
    "\\end{pmatrix}^T\\qquad$,$ v  = \\begin{pmatrix}\n",
    "1&2&3&4\\\\\n",
    "\\end{pmatrix}^T\\qquad$<br>\n",
    "(c)$ u  = \\begin{pmatrix}\n",
    "\\pi&\\sqrt2&0&1\\\\\n",
    "\\end{pmatrix}^T\\qquad$,$ v  = \\begin{pmatrix}\n",
    "1/\\pi&\\sqrt2&-1&1\\\\\n",
    "\\end{pmatrix}^T\\qquad$"
   ]
  },
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   "id": "3b73a002",
   "metadata": {},
   "source": [
    "<b>(a)</b><br> \n",
    "$ u  = \\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "-8\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix},  v  = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "2\\\\\n",
    "-5\\\\\n",
    "-3\\\\\n",
    "\\end{pmatrix}$<br> \n",
    "\n",
    "$u \\cdot v = 2(-1) + 3(2) + (-8)(-5) + 1(-3) = 41$\n",
    "\n",
    "\n",
    "$ \\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{2^2 + 3^2 + (-8)^2 + 1^2} \\cdot \\sqrt{(-1)^2 + 2^2 + (-5)^2 + (-3)^2} = \\sqrt{78} \\cdot \\sqrt{39}$\n",
    "\n",
    "$cos(\\theta) = \\frac{41}{\\sqrt{78} \\cdot \\sqrt{39}} =  0.7433$ <br>\n",
    "$ \\Rightarrow \\theta = cos^{-1}(0.7433)$\n",
    "\n",
    "<b>(b)</b> <br>\n",
    "$ u  = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "-3\\\\\n",
    "-1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix},  v  = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}$<br> \n",
    "\n",
    "$u \\cdot v = (-2)1 + (-3)2 + (-1)3 + (-1)4 = -15$\n",
    "\n",
    "\n",
    "$ \\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{(-2)^2 + (-3)^2 + (-1)^2 + (-1)^2} \\cdot \\sqrt{1^2 + 2^2 + 3^2 + 4^2} = \\sqrt{15} \\cdot \\sqrt{30}$\n",
    "\n",
    "$cos(\\theta) = \\frac{-15}{\\sqrt{15}\\cdot \\sqrt{30}} = -\\frac{1}{\\sqrt2}$\n",
    "\n",
    "$\\Rightarrow \\theta = 135^o$\n",
    "\n",
    "<b>(c)</b> <br>\n",
    "\n",
    "$ u  = \\begin{pmatrix}\n",
    "\\pi\\\\\n",
    "\\sqrt2\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix},  v  = \\begin{pmatrix}\n",
    "\\frac{1}{\\pi}\\\\\n",
    "\\sqrt{-2}\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$<br> \n",
    "\n",
    "$u \\cdot v = \\pi(\\frac{1}{\\pi}) + \\sqrt2 + \\sqrt2 + 0(-1) + 1(1) = 4$ \n",
    "\n",
    "$\\lVert u \\rVert \\cdot \\lVert v \\rVert = \\sqrt{(\\pi)^2 + (\\sqrt2)^2 + 0^2 + 1^2} \\cdot \\sqrt{(\\frac{1}{\\pi})^2 + \\sqrt2^2 + (-1)^2 + 1^2}  = 3.5874 \\cdot 2.0251$\n",
    "\n",
    "$ cos\\theta = \\frac{4}{3.5874 \\times 2.0251} =  0.5505$\n",
    "\n",
    "\n",
    "$ \\Rightarrow \\theta = cos^{-1}0.5505$"
   ]
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   "source": [
    "---------------------------------------------------------"
   ]
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   "id": "041aed26",
   "metadata": {},
   "source": [
    "<b> Question 4</b>\n",
    "\n",
    "Determine the value of $k$ so that the following vectors are orthogonal to each other:\n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "5\\\\\n",
    "k\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "-3\\\\\n",
    "2\\\\\n",
    "7\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "(b) $u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "-1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "3\\\\\n",
    "1\\\\\n",
    "k\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "(c) $u = \\begin{pmatrix}\n",
    "0\\\\\n",
    "-k\\\\\n",
    "\\sqrt2\\\\\n",
    "\\end{pmatrix}$,$v = \\begin{pmatrix}\n",
    "-7\\\\\n",
    "5\\\\\n",
    "k\\\\\n",
    "\\end{pmatrix}\\qquad$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0bc6984a",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "<b>(a)</b> <br>\n",
    "For vectors to be orthogonal, $u \\cdot v = 0$\n",
    "So, $(-1)(-3) + 5(2) + k(7) = 0$<br>\n",
    "$\\Rightarrow 3 + 10 + 7k = 0$\n",
    "\n",
    "$\\Rightarrow k = \\frac{-13}{7}$\n",
    "\n",
    "<b>(b)</b> <br>\n",
    "For vectors to be orthogonal, $u \\cdot v = 0$\n",
    "So, $2(3) + (-1)(1) + 3(k) = 0$<br>\n",
    "$\\Rightarrow 6 -1 + 3k = 0$\n",
    "\n",
    "$\\Rightarrow k = \\frac{-5}{3}$\n",
    "\n",
    "<b>(c)</b> <br>\n",
    "For vectors to be orthogonal, $u \\cdot v = 0$\n",
    "So, $0(-7) + (-k)(5) + \\sqrt2(k) = 0$<br>\n",
    "$\\Rightarrow 0 -5k + \\sqrt2k = 0$<br>\n",
    "$\\Rightarrow (\\sqrt2 - 5)k = 0$\n",
    "\n",
    "$\\Rightarrow k =0$"
   ]
  },
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   "source": [
    "--------------------------------------"
   ]
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   "metadata": {},
   "source": [
    "<b> Question 5 </b> <br>\n",
    "\n",
    "Determine the unit vector $\\hat u$ for each of the following vectors(Normalize the vectors)\n",
    "\n",
    "(a)$ u = \\begin{pmatrix}\n",
    "2&3\\\\\n",
    "\\end{pmatrix}^T\\qquad$(b)$ u = \\begin{pmatrix}\n",
    "1&2&3\\\\\n",
    "\\end{pmatrix}^T\\qquad$(c)$ u = \\begin{pmatrix}\n",
    "1/2&-1/2&1/4\\\\\n",
    "\\end{pmatrix}^T\\qquad$<br>\n",
    "(d)$ u = \\begin{pmatrix}\n",
    "\\sqrt2&2&-\\sqrt2&\\sqrt2\\\\\n",
    "\\end{pmatrix}^T\\qquad$(e)$ u = \\begin{pmatrix}\n",
    "-\\pi/5&\\pi&-\\pi&\\pi/10&0\\\\\n",
    "\\end{pmatrix}^T\\qquad$"
   ]
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   "id": "54267a35",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "<b>(a)</b> <br>\n",
    "$u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$ and we know that $\\hat u = \\frac{1}{\\lVert u \\rVert} \\cdot u$\n",
    "\n",
    "Here, $\\lVert u \\rVert =\\lVert \\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\rVert = \\sqrt{2^2 + 3^2} = \\sqrt{13}$\n",
    "\n",
    "$\\Rightarrow \\hat u = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "<b>(b)</b> <br>\n",
    "$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Here, $\\lVert u \\rVert =\\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\rVert = \\sqrt{1^2 2^2 + 3^2} = \\sqrt{14}$\n",
    "\n",
    "\n",
    "$\\Rightarrow \\hat u = \\frac{1}{\\sqrt{14}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(c)</b> <br>\n",
    "$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Here, $\\lVert u \\rVert =\\lVert \\begin{pmatrix}\n",
    "1/2\\\\\n",
    "-1/2\\\\\n",
    "1/4\\\\\n",
    "\\end{pmatrix} \\rVert = \\sqrt{(1/2)^2 (-1/2)^2 + (1/4)^2} = \\sqrt{1/4 + 1/4 + 16} = \\sqrt{\\frac{9}{16}} = \\frac{3}{4}$\n",
    "\n",
    "\n",
    "$\\Rightarrow \\hat u = \\frac{1}{3/4}\\begin{pmatrix}\n",
    "1/2\\\\\n",
    "-1/2\\\\\n",
    "1/4\\\\\n",
    "\\end{pmatrix} = \\frac{4}{3} \\cdot \\frac{1}{4}\\begin{pmatrix}\n",
    "2\\\\\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\hat u = \\frac{1}{3} \\begin{pmatrix}\n",
    "2\\\\\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(d)</b> <br>\n",
    "\n",
    "Here, $ u = \\begin{pmatrix}\n",
    "\\sqrt2\\\\\n",
    "2\\\\\n",
    "-\\sqrt2\\\\\n",
    "\\sqrt2\\\\\n",
    "\\end{pmatrix} = \\sqrt2 \\cdot \\begin{pmatrix}\n",
    "1\\\\\n",
    "\\sqrt2\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "So, $\\lVert u \\rVert = \\sqrt2 \\cdot \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "\\sqrt2\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\rVert = \\sqrt2 \\sqrt{1^2 + \\sqrt2^2 + (-1)^2 + 1^2} = \\sqrt2 \\cdot \\sqrt5$\n",
    "\n",
    "Therefore, $\\hat u = \\frac{1}{\\sqrt2 \\cdot \\sqrt5} \\cdot \\sqrt2 \\begin{pmatrix}\n",
    "1\\\\\n",
    "\\sqrt2\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = \\frac{1}{\\sqrt5} \\begin{pmatrix}\n",
    "1\\\\\n",
    "\\sqrt2\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(e)</b> <br>\n",
    "Here, \n",
    "\n",
    "$ u = \\begin{pmatrix}\n",
    "-(\\pi/5)\\\\\n",
    "\\pi\\\\\n",
    "-\\pi\\\\\n",
    "\\pi/10\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} = \\frac{\\pi}{10}\\begin{pmatrix}\n",
    "-2\\\\\n",
    "10\\\\\n",
    "-10\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "So, $\\lVert u \\rVert = \\frac{\\pi}{10} \\lVert \\begin{pmatrix}\n",
    "-2\\\\\n",
    "10\\\\\n",
    "-10\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} \\rVert = \\frac{\\pi}{10}\\sqrt{(-2)^2 + 10^2 + (-10)^2 + 1^2 + 0^2}$\n",
    "\n",
    "$ \\Rightarrow \\lVert u \\rVert = \\frac{\\pi}{10} \\sqrt{205}$\n",
    "\n",
    "Therefore, as $\\hat u = \\frac{1}{\\lVert u \\rVert} \\cdot u$\n",
    "\n",
    "We have, $\\hat u = \\frac{1}{\\frac{\\pi}{10}\\sqrt{205}} \\cdot \\frac{\\pi}{10}\\begin{pmatrix}\n",
    "-2\\\\\n",
    "10\\\\\n",
    "-10\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Hence, $\\hat u = \\frac{1}{\\sqrt{205}}\\begin{pmatrix}\n",
    "-2\\\\\n",
    "10\\\\\n",
    "-10\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$"
   ]
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    "----------------------------------------------------------------------------------"
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   "source": [
    "<b> Question 6</b> <br>\n",
    "\n",
    "Determine the value(s) of $k$ so that $\\hat u = \\begin{pmatrix}\n",
    "1/\\sqrt2\\\\\n",
    "1/2\\\\\n",
    "k\\\\\n",
    "\\end{pmatrix}$ is a unit vector"
   ]
  },
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   "id": "3fc5a703",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br> \n",
    "\n",
    "We know, $\\lVert \\hat u \\rVert = 1$ \n",
    "\n",
    "$\\Rightarrow\\quad \\lVert \\begin{pmatrix}\n",
    "(1/\\sqrt{2})\\\\\n",
    "1/2\\\\\n",
    "k\\\\\n",
    "\\end{pmatrix} \\rVert = \\sqrt{(\\frac{1}{\\sqrt2})^2 + (\\frac{1}{2})^2 + k^2}$\n",
    "\n",
    "$\\Rightarrow\\quad \\frac{1}{2} + \\frac{1}{4} + k^2 = 1$ <br>\n",
    "$\\Rightarrow\\quad \\frac{3}{4} = k^2 = 1$ <br>\n",
    "$\\Rightarrow\\quad K^2 = 1 - \\frac{3}{4} = \\frac{1}{4}$ <br>\n",
    "\n",
    "$\\Rightarrow\\quad k = \\pm \\frac{1}{2}$"
   ]
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   "source": [
    "----------------------------------------------------------------------------------"
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   "source": [
    "<b> Question 7</b><br>\n",
    "\n",
    "(a)Show that $u = \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "sin(\\theta)\\\\\n",
    "\\end{pmatrix}$ is a unit vector.\n",
    "\n",
    "(b). Plot this vector $u$ in $\\mathbb{R}^2$ for $\\theta = \\frac{\\pi}{4}$\n",
    "\n",
    "(c) Let $v =  \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "-sin(\\theta)\\\\\n",
    "\\end{pmatrix}$ be a vector in $\\mathbb{R}^2$. On the same axes plot $v$ for $\\theta = \\frac{\\pi}{4}$\n",
    "\n",
    "(d). DEtermine the angle between the vectors $u$ and $v$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d921181b",
   "metadata": {},
   "source": [
    "<b> Solution </b> <br> \n",
    "<b>(a). </b>\n",
    "For unit vector, $\\lVert \\hat u \\rVert = 1$ \n",
    "\n",
    "we have, $u = \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "sin(\\theta)\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\lVert \\hat u \\rVert  = \\lVert \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "sin(\\theta)\\\\\n",
    "\\end{pmatrix} \\rVert = \\sqrt{cos^{2}\\theta + sin^{2}\\theta} = 1$\n",
    "\n",
    "Therefore,  $u = \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "sin(\\theta)\\\\\n",
    "\\end{pmatrix}$ is a unit vector.\n",
    "\n",
    "<b>(b)</b> I will learn Latex plot drawing later<br>\n",
    "<b>(c)</b> I will learn Latex plot drawing later<br>\n",
    "\n",
    "\n",
    "<b>(d)</b> <br>\n",
    "\n",
    "We know, $cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$\n",
    "\n",
    "Here, we have\n",
    "$u = \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "sin(\\theta)\\\\\n",
    "\\end{pmatrix}$ and $v =  \\begin{pmatrix}\n",
    "cos(\\theta)\\\\\n",
    "-sin(\\theta)\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "So, as we have to use $\\theta = \\frac{\\pi}{4}$<br>\n",
    "$u  \\cdot v = \\begin{pmatrix}\n",
    "cos(\\frac{\\pi}{4})\\\\\n",
    "sin(\\frac{\\pi}{4})\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "cos(\\frac{\\pi}{4})\\\\\n",
    "-sin(\\frac{\\pi}{4})\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "\\frac{1}{\\sqrt2}\\\\\n",
    "\\frac{1}{\\sqrt2}\\\\\n",
    "\\end{pmatrix}\\cdot\\begin{pmatrix}\n",
    "\\frac{1}{\\sqrt2}\\\\\n",
    "-\\frac{1}{\\sqrt2}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$u  \\cdot v = \\frac{1}{\\sqrt2} \\times \\frac{1}{\\sqrt2} + \\frac{1}{\\sqrt2} \\times-(\\frac{1}{\\sqrt2}) = \\frac{1}{2} - \\frac{1}{2} = 0$\n",
    "\n",
    "Therefore, $u$ and $v$ are orthogonal. Angle would be $90^o$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9122395c",
   "metadata": {},
   "source": [
    "----------------------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7dfecc8e",
   "metadata": {},
   "source": [
    "<b> Question 8 </b> <br>\n",
    "\n",
    "Show that vectors $u = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$ and $v = \\begin{pmatrix}\n",
    "-b\\\\\n",
    "a\\\\\n",
    "\\end{pmatrix}$ in $\\mathbb{R}^2$ are orthogonal."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d18abdd5",
   "metadata": {},
   "source": [
    "<b>Solution </b> <br> \n",
    "For orthogonal, we need $u \\cdot v = 0$ \n",
    "\n",
    "Here, $u \\cdot v = u = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-b\\\\\n",
    "a\\\\\n",
    "\\end{pmatrix} = a(-b) + a(b) = -ab + ab = 0$\n",
    "\n",
    "Hence, proved!"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9a3a92a0",
   "metadata": {},
   "source": [
    "----------------------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c45fc66a",
   "metadata": {},
   "source": [
    "<b> Question 9</b> <br>\n",
    "\n",
    "Let vectors $u = \\begin{pmatrix}\n",
    "cos(A)\\\\\n",
    "sin(A)\\\\\n",
    "\\end{pmatrix}$ and  $v =\\begin{pmatrix}\n",
    "cos(B)\\\\\n",
    "sin(B)\\\\\n",
    "\\end{pmatrix}$ be in $\\mathbb{R}^2$. Show that $u \\cdot v = cos(A - B)$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a39bd59e",
   "metadata": {},
   "source": [
    "<b> Solution </b> <br> \n",
    "\n",
    "Here, $u \\cdot v = \\begin{pmatrix}\n",
    "cos(A)\\\\\n",
    "sin(A)\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "cos(B)\\\\\n",
    "sin(B)\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$ = cos(A)cos(B) + sin(A)sin(B)$\n",
    "\n",
    "\n",
    "As, cos(A)cos(B) + sin(A)sin(B) = cos(A -B) <br>\n",
    "\n",
    "We have, $u \\cdot v = cos(A-B)$\n",
    "\n",
    "Hence, proved!\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ff38f922",
   "metadata": {},
   "source": [
    "----------------------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "148b91eb",
   "metadata": {},
   "source": [
    "<b>Question 10</b><br>\n",
    "\n",
    "Prove that the angle between a non-zero vector $u$ and $-u$ in any n-space, $\\mathbb{R}^n$, is $\\pi$ radians or 180 degree"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "75636213",
   "metadata": {},
   "source": [
    "<b> Solution </b> <br> \n",
    "\n",
    "We know, <br>\n",
    "\n",
    "$cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$\n",
    "\n",
    "Here, $u = u$ and $v = -u$\n",
    "\n",
    "Therefore, $u \\cdot v = u \\cdot (-u)$ <br>\n",
    "$= u \\cdot (-1)u$<br>\n",
    "$= (-1)(u \\cdot u)$<br>\n",
    "$= (-1)\\lVert u \\rVert^2$ as$\\qquad  \\sqrt{u \\cdot u} = \\lVert u \\rVert$\n",
    "\n",
    "Next, $\\lVert v \\rVert$ <br>\n",
    "\n",
    "$= \\lVert -u \\rVert$\n",
    "\n",
    "we know that, $\\lVert ku \\rVert = \\lvert k \\rvert \\lVert u \\rVert$\n",
    "\n",
    "Therefore, $\\lVert -u \\rVert = \\lVert (-1)u \\rVert$<br>\n",
    "$\\qquad\\qquad\\qquad =  \\lvert -1 \\rvert \\lVert u \\rVert$<br>\n",
    "$\\Rightarrow \\quad \\qquad \\lVert v \\rVert = 1\\lVert u \\rVert = \\lVert u \\rVert$\n",
    "\n",
    "Now, <br>\n",
    "\n",
    "$cos(\\theta) = \\frac{u \\cdot v}{\\lVert u \\rVert \\cdot \\lVert v \\rVert}$\n",
    "\n",
    "$\\Rightarrow \\qquad cos(\\theta) = \\frac{(-1)\\lVert u \\rVert^2}{\\lVert u \\rVert \\cdot \\lVert u \\rVert}$\n",
    "\n",
    "$\\Rightarrow \\qquad cos(\\theta) = \\frac{(-1)\\lVert u \\rVert^2}{\\lVert u \\rVert^2} = -1$\n",
    "\n",
    "$\\Rightarrow \\qquad \\theta = cos^{-1}(-1) = 180^o$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2243b112",
   "metadata": {},
   "source": [
    "----------------------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "be50cdfe",
   "metadata": {},
   "source": [
    "<b> Question 11</b> <br>\n",
    "\n",
    "Find a vector which is orthogonal to $u = \\begin{pmatrix}\n",
    "1&1&1\\\\\n",
    "\\end{pmatrix}^T$. Determine all the vectors orthogonal to $u$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ed63af14",
   "metadata": {},
   "source": [
    "<b> Solution</b> <br>\n",
    "\n",
    "Let vector orthogonal to $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$ as $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "We know that, \n",
    "\n",
    "For orthogonal $ u \\cdot v = 0$ \n",
    "\n",
    "Therefore, $\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow x + y + z = 0$\n",
    "\n",
    "$\\Rightarrow x = -y -z$\n",
    "\n",
    "\n",
    "Here, we can have infinte solution to equation, but as we have 3 unknowns, no of free variables are $3-1 = 2$<br>\n",
    "Therefore, <br>\n",
    "\n",
    "if $y = a$ and $z = b$\n",
    "\n",
    "then, $x = -a -b$\n",
    "\n",
    "Hence, orthogonal vector $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "-a -b\\\\\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "81332c66",
   "metadata": {},
   "source": [
    "----------------------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "12e9ef04",
   "metadata": {},
   "source": [
    "<b> Question 12</b> <br>\n",
    "\n",
    "Find the shortest distance, correct to 2dp, between the vectors and the corresponding hyperplanes:\n",
    "\n",
    "(a)$\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "\\end{pmatrix}^T$ and $ y = x + 1$ <br>\n",
    "(b)$\\begin{pmatrix}\n",
    "0.5&2\\\\\n",
    "\\end{pmatrix}^T$ and $ y = 2x - 1$<br>\n",
    "(c)$\\begin{pmatrix}\n",
    "-1&-3&3\\\\\n",
    "\\end{pmatrix}^T$ and $ 2x + y -z = 7$ <br>\n",
    "(a)$\\begin{pmatrix}\n",
    "1&2&3&4\\\\\n",
    "\\end{pmatrix}^T$ and $ x + 2y + z + w = 10$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0b5789aa",
   "metadata": {},
   "source": [
    "<b> Solution</b> <br> \n",
    "\n",
    "(a) Here, equation is $y = x +1$ \n",
    "\n",
    "\n",
    "$\\Rightarrow x - y +1 = 0$\n",
    "\n",
    "In dot production form, it can be written as :\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} + 1 = 0\\qquad$ \n",
    "\n",
    "This is in $ v \\cdot x + c = 0$ form and we have $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $\\frac{\\lvert u \\cdot v + c \\rvert}{\\lVert v \\rVert} =  \\frac{\\lvert \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} + 1\\rvert}{\\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\rVert}$\n",
    "\n",
    "$ = \\frac{1-1+1}{\\sqrt{1^2 + 1^2}} = \\frac{1}{\\sqrt2} = 0.71$\n",
    "\n",
    "Hence, shortest distance = 0.71\n",
    "\n",
    "<b> Solution (b)</b><br>\n",
    "\n",
    "Here, equation is $y = 2x -1$ \n",
    "\n",
    "\n",
    "$\\Rightarrow 2x - y -1 = 0$\n",
    "\n",
    "In dot production form, it can be written as :\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "2\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} + (-1) = 0\\qquad$ \n",
    "\n",
    "This is in $ v \\cdot x + c = 0$ form and we have $u = \\begin{pmatrix}\n",
    "0.5\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $\\frac{\\lvert u \\cdot v + c \\rvert}{\\lVert v \\rVert} =  \\frac{\\lvert \\begin{pmatrix}\n",
    "0.5\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "2\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} + (-1)\\rvert}{\\lVert \\begin{pmatrix}\n",
    "2\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}\\rVert}$\n",
    "\n",
    "$ = \\frac{0.5(2) + 2(-1) +(-1)}{\\sqrt{2^2 + (-1)^2}} = \\frac{2}{\\sqrt5} = 0.89$\n",
    "\n",
    "Hence, shortest distance = 0.89\n",
    "\n",
    "\n",
    "\n",
    "<b> Solution (c)</b><br>\n",
    "\n",
    "Here, equation is $2x + y - z = 7$ \n",
    "\n",
    "\n",
    "$\\Rightarrow 2x + y -z -7 = 0$\n",
    "\n",
    "In dot production form, it can be written as :\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "2\\\\\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} + (-1) = 0\\qquad$ \n",
    "\n",
    "This is in $ v \\cdot x + c = 0$ form and we have $u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "-3\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $\\frac{\\lvert u \\cdot v + c \\rvert}{\\lVert v \\rVert} =  \\frac{\\lvert \\begin{pmatrix}\n",
    "-1\\\\\n",
    "-3\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "2\\\\\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} + (-7)\\rvert}{\\lVert \\begin{pmatrix}\n",
    "2\\\\\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}\\rVert}$\n",
    "\n",
    "$ = \\frac{\\lvert-1(2) + (-3)1 +3(-1) -7\\rvert}{\\sqrt{2^2 + 1^2 + (-1)^2}} = \\frac{\\lvert -2-3-3-7 \\rvert}{\\sqrt6} = \\frac{15}{\\sqrt6} = 6.12$\n",
    "\n",
    "Hence, shortest distance = 6.12\n",
    "\n",
    "\n",
    "\n",
    "<b> Solution (d)</b><br>\n",
    "\n",
    "Here, equation is $x + 2y + z+ w  = 10$ \n",
    "\n",
    "\n",
    "$\\Rightarrow x + 2y +z +w -10 = 0$\n",
    "\n",
    "In dot production form, it can be written as :\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "1\\\\\n",
    "1\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "w\\\\\n",
    "\\end{pmatrix} + (-10) = 0\\qquad$ \n",
    "\n",
    "This is in $ v \\cdot x + c = 0$ form and we have $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $\\frac{\\lvert u \\cdot v + c \\rvert}{\\lVert v \\rVert} =  \\frac{\\lvert \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + (-10)\\rvert}{\\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\rVert}$\n",
    "\n",
    "$ = \\frac{\\lvert1(1) + (2)2 +3(1) + 4(1) -10 \\rvert}{\\sqrt{1^2 + 2^2 + 1^2 + 1^2}} = \\frac{\\lvert 1+ 4+ 3+ 4 -10 \\rvert}{\\sqrt7} = \\frac{2}{\\sqrt{7}} = 0.76$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "be585978",
   "metadata": {},
   "outputs": [],
   "source": []
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