Vi bestämmer först skärningslinjen mellan planen. Vi får

\left\{ \begin{array}{l} 3x + 2y - 5z = 5 \\ 4x - y - 3z = 3 \end{array} \right. \Leftrightarrow \left( \begin{array}{ccc} 3 & 2 & -5|5 \\ 4 & -1 & -3|3 \end{array} \right) \Leftrightarrow \left( \begin{array}{ccc} 3 & 2 & -5|5 \\ 12 & -3 & -9|9 \end{array} \right) \Leftrightarrow \left( \begin{array}{ccc} 3 & 2 & -5|5 \\ 0 & -11 & 11|-11 \end{array} \right) \Leftrightarrow \left( \begin{array}{ccc} 3 & 2 & -5|5 \\ 0 & -1 & 1|-1 \end{array} \right) \Leftrightarrow \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) + s \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right).

Vi ritar en figur.

I) $P$ : $s$ ortogonala projektion $Q$ på linjen sökes.

Vi konstaterar att origo ej ligger på linjen (sätt $s = 0$ i linjens ekvation). Vi bildar vektorn

\boldsymbol{u} = \overline{P_0P} = \left( \begin{array}{c} 2 \\ 1 \\ 3 \end{array} \right) - \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} 1 \\ 0 \\ 3 \end{array} \right)

Projektionsformeln ger

\boldsymbol{u}_{\parallel \nu} = \overline{P_0Q} = \frac{ \left( \begin{array}{c} 1 \\ 0 \\ 3 \end{array} \right) \cdot \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) }{3} \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) = \frac{4}{3} \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right)

vilket ger

\overline{OQ} = \overline{OP_0} + \overline{P_0Q} = \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) + \frac{4}{3} \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) = \left( \begin{array}{c} 7/3 \\ 7/3 \\ 4/3 \end{array} \right)

som ger projektionspunkten $\left( \frac{7}{3}, \frac{7}{3}, \frac{4}{3} \right)$ .