\(% 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}}\)
The line through \(A\) perpendicular to \(BC\) is \[ l_1: \vect {r} = \vect {a} + \lambda \vect {u}, \lambda \in \RR . \]
The line through \(B\) perpendicular to \(CA\) is \[ l_2: \vect {r} = \vect {b} + \mu \vect {v}, \mu \in \RR . \]
Since \(P\) is the intersection of \(l_1\) and \(l_2\), we must have \[ \vect {a} + \lambda \vect {u} = \vect {b} + \mu \vect {v}, \] and hence solving for \(\vect {v}\) we have \[ \vect {v} = \frac {1}{\mu } \left (\vect {a} + \lambda \vect {u} - \vect {b}\right ). \]
Since \(\vect {v}\) is perpendicular to \(CA\), we must have \(\vect {v} \cdot (\vect {a} - \vect {c}) = 0\), and hence \begin {align*} & \phantom {\iff } \frac {1}{\mu } \left (\vect {a} + \lambda \vect {u} - \vect {b}\right ) \cdot (\vect {a} - \vect {c}) = 0 \\ & \iff (\vect {a} - \vect {b}) \cdot (\vect {a} - \vect {c}) + \lambda \vect {u} \cdot (\vect {a} - \vect {c}) = 0 \\ & \iff \lambda = -\frac {(\vect {a} - \vect {b}) \cdot (\vect {a} - \vect {c})}{\vect {u} \cdot (\vect {a} - \vect {c})}. \end {align*}
Hence, the position vector of \(P\), \(\vect {p}\), must satisfy that \[ \vect {p} = \vect {a} + \lambda \vect {u} = \vect {a} - \frac {(\vect {a} - \vect {b}) \cdot (\vect {a} - \vect {c})}{\vect {u} \cdot (\vect {a} - \vect {c})} \vect {u}. \]
\(CP\) is perpendicular to \(AB\) if and only if \((\vect {p} - \vect {c}) \cdot (\vect {b} - \vect {a}) = 0\). We notice \begin {align*} (\vect {p} - \vect {c}) \cdot (\vect {b} - \vect {a}) & = \left (\vect {a} + \lambda \vect {u} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) \\ & = \left (\vect {a} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) + \lambda \vect {u} \left (\vect {b} - \vect {a}\right ). \end {align*}
Since \(\vect {u}\) is perpendicular to \(BC\), we must have \(\vect {u} \cdot (\vect {c} - \vect {b})\), and hence \(\vect {u} \cdot \vect {c} = \vect {u} \cdot \vect {b}\). Hence, \begin {align*} (\vect {p} - \vect {c}) \cdot (\vect {b} - \vect {a}) & = \left (\vect {a} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) + \lambda \vect {u} \cdot \left (\vect {b} - \vect {a}\right ) \\ & = \left (\vect {a} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) + \lambda \vect {u} \cdot \left (\vect {c} - \vect {a}\right ) \\ & = \left (\vect {a} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) - \frac {(\vect {a} - \vect {b}) \cdot (\vect {a} - \vect {c})}{\vect {u} \cdot (\vect {a} - \vect {c})} \vect {u} \cdot \left (\vect {c} - \vect {a}\right ) \\ & = \left (\vect {a} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) + (\vect {a} - \vect {b}) \cdot (\vect {a} - \vect {c}) \\ & = \left (\vect {a} - \vect {c}\right ) \cdot (\vect {b} - \vect {a}) - (\vect {a} - \vect {c}) \cdot (\vect {b} - \vect {a}) \\ & = 0, \end {align*}
and hence \(CP\) is perpendicular to \(AB\).