下の図の三角柱OAB-CDEにおいて，$\overrightarrow{\mathrm{OA}}=\overrightarrow{a},\ \overrightarrow{\mathrm{OB}}=\overrightarrow{b},\ \overrightarrow{\mathrm{OC}}=\overrightarrow{c}$とおき， \begin{align} & |\overrightarrow{a}|=\sqrt{3},\quad |\overrightarrow{b}|=\sqrt{5},\quad |\overrightarrow{c}|=4 \nonumber \\ & \overrightarrow{a} \cdot \overrightarrow{b} =1,\quad \overrightarrow{a} \cdot \overrightarrow{c}=\overrightarrow{b} \cdot \overrightarrow{c}=0 \nonumber \end{align} とする．辺AD，BE上にそれぞれ点P，Qをとり，$\text{AP}=s,\ \text{BQ}=t$とおく．
(1) $\overrightarrow{\mathrm{OP}},\ \overrightarrow{\mathrm{PQ}}$を$\overrightarrow{a},\ \overrightarrow{b},\ \overrightarrow{c}$および$s,\ t$を用いて表せ．
(2) $\text{OP} \perp \text{PQ}$となるとき，$t$を$s$を用いて表せ．
(3) $\triangle$OPQが$\text{OP}=\text{PQ}$の直角二等辺三角形となるように，$s,\ t$の値を定めよ．
\setlength\unitlength{1truecm} \begin{center} \begin{picture}(4,6.5)(0,0) \multiput(0.5,2)(0.20,0){12}{\line(1,0){0.1}} \put(0.5,2){\line(0,1){4}} \put(0.5,2){\line(1,-2){0.8}} \put(1.3,0.4){\line(1,1){1.58}} \put(0.5,6){\line(1,-2){0.8}} \put(0.5,6){\line(1,0){2.4}} \put(2.9,6){\line(0,-1){4}} \put(2.9,6){\line(-1,-1){1.58}} \put(1.3,0.4){\line(0,1){4}} \put(0.1,1.8){O} \put(1.4,0.2){A} \put(3,1.8){B} \put(0.1,5.8){C} \put(1.4,4.2){D} \put(3,5.8){E} \end{picture} \end{center}