Snijlijn van twee vlakken

In een rechthoekig assenstelsel OXYZ zijn gegeven de punten A(4,0,0), B(4,8,0), C (0,8,0) en T (0,0,8). Op AB ligt punt D(4,6,0).

• Bepaal een vectorvoorstelling van de snijlijn van vlak TDO en vlak ACT.

Uitwerking 

$\begin{array}{l} TDO:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + \mu \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ ACT:\left( {\begin{array}{*{20}c} 4 \\ 0 \\ 0 \\ \end{array}} \right) + \rho \left( {\begin{array}{*{20}c} { - 1} \\ 2 \\ 0 \\ \end{array}} \right) + \tau \left( {\begin{array}{*{20}c} { - 1} \\ 0 \\ 2 \\ \end{array}} \right) \\ \left\{ \begin{array}{l} 2\lambda = 4 - \rho - \tau \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\lambda = 8 - 2\rho - 2\tau \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\lambda = 8 - 3\lambda - \mu \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \mu = - 7\lambda + 8 \\ l:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + \left( { - 7\lambda + 8} \right)\left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ l:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + - 7\lambda \cdot \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) + 8 \cdot \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ l:\left( {\begin{array}{*{20}c} 0 \\ 0 \\ 8 \\ \end{array}} \right) + \lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ { - 7} \\ \end{array}} \right) \\ \end{array}$

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• Examenvraag 82-83 (4)