\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

[next] [prev] [prev-tail] [tail] [up]

2013.3.3 Question 3

Since \(\vect {p}_1 + \vect {p}_2 + \vect {p}_3 + \vect {4}_4 = \vect {0}\), we must have \begin {align*} 0 & = \vect {0} \cdot \vect {0} \\ & = \left (\vect {p}_1 + \vect {p}_2 + \vect {p}_3 + \vect {4}_4\right ) \cdot \left (\vect {p}_1 + \vect {p}_2 + \vect {p}_3 + \vect {4}_4\right ) \\ & = \sum _{i = 1}^{4} \vect {p}_i \cdot \vect {p}_i + 2 \sum _{i = 1}^{3} \sum _{j = i + 1}^{4} \vect {p}_i \cdot \vect {p}_j. \end {align*}

Since \(P_i\) are on the unit sphere, we must have \(\vect {p}_i \cdot \vect {p}_i = 1\). By symmetry, for all \(i \neq j\), \[ \vect {p}_i \cdot \vect {p}_j \] must be some real constant, say \(k\).

Hence, \[ 0 = 4 \cdot 1 + 2 \cdot 6 \cdot k, \] which solves to \[ k = -\frac {1}{3}, \] as desired.

  1. We have \begin {align*} \sum _{i = 1}^{4} (X P_i)^2 & = \sum _{i = 1}^{4} \left (\vect {p}_i - \vect {x}\right ) \cdot \left (\vect {p}_i - \vect {x}\right ) \\ & = \sum _{i = 1}^{4} \left (\vect {p_i} \cdot \vect {p_i} - 2 \vect {x} \cdot \vect {p}_i + \vect {x} \cdot \vect {x}\right ) \\ & = \sum _{i = 1}^{4} \vect {p_i} \cdot \vect {p_i} - 2 \vect {x} \cdot \sum _{i = 1}^{4} + 4 \cdot \vect {x} \cdot \vect {x} \\ & = \sum _{i = 1}^{4} 1 - 2 \vect {x} \cdot \vect {0} + 4 \cdot 1 \\ & = 4 - 0 + 4 \\ & = 8. \end {align*}

  2. Since \(P_1 (0, 0, 1)\) and \(P_2 (a, 0, b)\), we must have \[ \vect {p}_1 = \begin {pmatrix} 0 \\ 0 \\ 1 \end {pmatrix}, \vect {p}_2 = \begin {pmatrix} a \\ 0 \\ b \end {pmatrix}, \] and hence \[ \vect {p}_1 \cdot \vect {p}_2 = 0 \cdot a + 0 \cdot 0 + 1 \cdot b = b = - \frac {1}{3}. \]

    We must have \[ \abs *{\vect {p}_2} = \sqrt {a^2 + 0^2 + b^2} = \sqrt {a^2 + b^2} = 1, \] which means \[ a = \frac {2\sqrt {2}}{3}, \] as desired.

    The \(z\)-component of \(\vect {p}_3\) and \(\vect {p}_4\) must also be \(-\frac {1}{3}\), due to the dot product with \(vect{p}_1\) being equal to the \(z\)-component must also be equal to \(-\frac {1}{3}\).

    Let \[ \vect {p}_3 = \begin {pmatrix} c \\ d \\ -\frac {1}{3} \end {pmatrix}, \] then from \(\sum _{i = 1}^{4} \vect {p}_i = \vect {0}\), we have \[ \vect {p}_4 = \begin {pmatrix} - c - \frac {2\sqrt {2}}{3} \\ -d \\ -\frac {1}{3} \end {pmatrix}. \]

    Since \(\vect {p}_3 \cdot \vect {p}_2 = -\frac {1}{3}\), we have \[ \frac {2\sqrt {2}}{3} \cdot c + 0 \cdot d + \left (-\frac {1}{3}\right ) \cdot \left (-\frac {1}{3}\right ) = -\frac {1}{3}, \] and hence \[ \frac {2\sqrt {2}}{3} c = -\frac {4}{9}, \] which means \[ 6\sqrt {2} c = -4, \] and hence \[ c = - \frac {4}{6\sqrt {2}} = - \frac {\sqrt {2}}{3}. \]

    Now, since \(\vect {p}_3 \cdot \vect {p}_4 = - \frac {1}{3}\), we have \[ c \cdot \left (- c - \frac {2\sqrt {2}}{3}\right ) + d \cdot (-d) + \left (-\frac {1}{3}\right ) \cdot \left (-\frac {1}{3}\right ) = - \frac {1}{3}. \]

    Therefore, \[ \left (-\frac {\sqrt {2}}{3}\right ) \cdot \left (- \frac {\sqrt {2}}{3}\right ) - d^2 = -\frac {4}{9}, \] and hence \[ d^2 = \frac {2}{3}, \] giving \[ d = \pm \frac {\sqrt {2}}{\sqrt {3}}. \]

    Hence, \[ P_3 \left (-\frac {\sqrt {2}}{3}, \pm \frac {\sqrt {2}}{\sqrt {3}}, -\frac {1}{3}\right ), P_4 \left (-\frac {\sqrt {2}}{\sqrt {3}}, \mp \frac {\sqrt {2}}{3}, -\frac {1}{3}\right ). \]

  3. We have \begin {align*} \sum _{i = 1}^{4} \left (X P_i\right )^4 & = \sum _{i = 1}^{4} \left [\left (\vect {p}_i - \vect {x}\right ) \cdot \left (\vect {p}_i - \vect {x}\right )\right ]^2 \\ & = \sum _{i = 1}^{4} \left (\vect {p}_i \cdot \vect {p}_i - 2 \vect {x} \cdot \vect {p}_i + \vect {x} \cdot \vect {x}\right )^2 \\ & = \sum _{i = 1}^{4} \left (1 + 1 - 2 \vect {x} \cdot \vect {p}_i\right )^2 \\ & = \sum _{i = 1}^{4} \left (2 - 2 \vect {x} \cdot \vect {p}_i\right )^2 \\ & = 4 \sum _{i = 1}^{4} \left (1 - \vect {x} \cdot \vect {p}_i\right )^2. \end {align*}

    Let \(X(x, y, z)\). We have \begin {align*} \sum _{i = 1}^{4} \left (X P_i\right )^4 & = 4 \sum _{i = 1}^{4} \left (1 - \vect {x} \cdot \vect {p}_i\right )^2 \\ & = 4 \left [ \left (1 - \begin {pmatrix} x \\ y \\ z \end {pmatrix} \cdot \begin {pmatrix} 0 \\ 0 \\ 1 \end {pmatrix} \right )^2 + \left (1 - \begin {pmatrix} x \\ y \\ z \end {pmatrix} \cdot \begin {pmatrix} \frac {2\sqrt {2}}{3} \\ 0 \\ -\frac {1}{3} \end {pmatrix} \right )^2 \right . \\ & \phantom {=} \left . + \left (1 - \begin {pmatrix} x \\ y \\ z \end {pmatrix} \cdot \begin {pmatrix} -\frac {\sqrt {2}}{3} \\ \frac {\sqrt {2}}{\sqrt {3}} \\ -\frac {1}{3} \end {pmatrix} \right )^2+ \left (1 - \begin {pmatrix} x \\ y \\ z \end {pmatrix} \cdot \begin {pmatrix} -\frac {\sqrt {2}}{3} \\ -\frac {\sqrt {2}}{\sqrt {3}} \\ -\frac {1}{3} \end {pmatrix} \right )^2\right ] \\ & = 4 \left [(1 - z)^2 + \left (1 - \frac {2\sqrt {2}}{3}x + \frac {1}{3}z\right )^2 \right . \\ & \phantom {=} \left . + \left (1 + \frac {\sqrt {2}}{3}x - \frac {\sqrt {2}}{\sqrt {3}}y + \frac {1}{3}z\right )^2 + \left (1 + \frac {\sqrt {2}}{3}x + \frac {\sqrt {2}}{\sqrt {3}}y + \frac {1}{3}z\right )^2 \right ] \\ & = 4\left (4 + \frac {4}{3} x^2 + \frac {4}{3}y^2 + \frac {4}{3}z^2\right ) \\ & = 4 \left [4 + \frac {4}{3}\right ] \\ & = 4 \cdot \frac {16}{3} \\ & = \frac {64}{3} \end {align*}

    is a constant, independent of the position of \(X\).

[next] [prev] [prev-tail] [front] [up]