【高校数学C】2-1-4 空間の位置ベクトル｜問題集

\(\displaystyle \left(\frac{1・0+2・3}{2+1},\frac{1・3+2・(-3)}{2+1},\frac{1・7+2・1}{2+1}\right)\)
\(\displaystyle =(2,-1,3)\)

\(\displaystyle \left(\frac{-2・0+3・3}{3-2},\frac{-2・3+3・(-3)}{3-2},\frac{-2・7+3・1}{3-2}\right)\)
\(\displaystyle =(9,-15,-11)\)

\(\displaystyle \left(\frac{3-6}{2},\frac{-3+2}{2},\frac{1-1}{2}\right)\)
\(\displaystyle =\left(-\frac{3}{2},-\frac{1}{2},0\right)\)

\(\displaystyle \left(\frac{0+3-6}{3},\frac{3-3+2}{3},\frac{7+1-1}{3}\right)\)
\(\displaystyle =\left(-1,\frac{2}{3},\frac{7}{3}\right)\)

\(\displaystyle \overrightarrow{OD}=\frac{\vec{a}+2\vec{b}}{2+1}\)
\(\displaystyle =\frac{1}{3}\vec{a}+\frac{2}{3}\vec{b}\)
\(\displaystyle \overrightarrow{OP}=\frac{3\overrightarrow{OD}+2\vec{c}}{2+3}\)
\(\displaystyle =\frac{3}{5}\overrightarrow{OD}+\frac{2}{5}\vec{c}\)
\(\displaystyle =\frac{3}{5}\left(\frac{1}{3}\vec{a}+\frac{2}{3}\vec{b}\right)+\frac{2}{5}\vec{c}\)
\(\displaystyle =\frac{1}{5}\vec{a}+\frac{2}{5}\vec{b}+\frac{2}{5}\vec{c}\)

\(\displaystyle \overrightarrow{AP}=\overrightarrow{OP}-\overrightarrow{OA}\)
\(\displaystyle =\frac{\vec{b}+\vec{c}}{2}-\vec{a}\)
\(\displaystyle =-\vec{a}+\frac{1}{2}\vec{b}+\frac{1}{2}\vec{c}\)
\(\displaystyle \overrightarrow{BQ}=\overrightarrow{OQ}-\overrightarrow{OB}\)
\(\displaystyle =\frac{3\vec{a}+2\vec{c}}{2+3}-\vec{b}\)
\(\displaystyle =\frac{3}{5}\vec{a}-\vec{b}+\frac{2}{5}\vec{c}\)

\(\displaystyle \overrightarrow{OP}=\frac{3}{4}\overrightarrow{OG}\)
\(\displaystyle =\frac{3}{4}\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)\)
\(\displaystyle =\frac{1}{4}\vec{a}+\frac{1}{4}\vec{b}+\frac{1}{4}\vec{c}\)

\(\displaystyle \overrightarrow{OG}=\frac{\overrightarrow{OO}+\overrightarrow{OA}+\overrightarrow{OB}}{3}\)
\(\displaystyle =\frac{1}{3}\vec{a}+\frac{1}{3}\vec{b}\)
\(\displaystyle \overrightarrow{OE}=\frac{-3\overrightarrow{OG}+5\overrightarrow{OD}}{5-3}\)
\(\displaystyle =-\frac{3}{2}\left(\frac{1}{3}\vec{a}+\frac{1}{3}\vec{b}\right)+\frac{5}{2}・\frac{4}{5}\vec{c}\)
\(\displaystyle =-\frac{1}{2}\vec{a}-\frac{1}{2}\vec{b}+2\vec{c}\)

\(\overrightarrow{AB}=(6,-2,2)\)
\(\overrightarrow{AC}=(-3,x-3,y-6)\)
\(\overrightarrow{AC}=k\overrightarrow{AB}\)より、
\((-3,x-3,y-6)=k(6,-2,2)\)
\(-3=6k,x-3=-2,y-6=2\)
これを解くと、
\(\displaystyle k=-\frac{1}{2},x=4,y=5\)
よって、
\(x=4,y=5\)

\(\overrightarrow{AB}=(6,0,3)\)
\(\overrightarrow{AC}=(5,-2,8)\)
\(\overrightarrow{AP}=(x+3,2,-2)\)
\(\overrightarrow{AP}=s\overrightarrow{AB}+t\overrightarrow{AC}\)より、
\((x+3,2,-2)=s(6,0,3)+t(5,-2,8)\)
\((x+3,2,-2)=(6s+5t,-2t,3s+8t)\)
\(x+3=6s+5t,2=-2t,-2=3s+8t\)
これを解くと、
\(t=-1,s=2,x=4\)
よって、
\(x=4\)

\(\displaystyle \overrightarrow{OR}=\frac{1}{2}\overrightarrow{OM}+\frac{1}{2}\overrightarrow{OQ}\)
\(\displaystyle =\frac{1}{2}・\frac{1}{2}\vec{a}+\frac{1}{2}\left(\frac{2\vec{b}+\vec{c}}{1+2}\right)\)
\(\displaystyle =\frac{1}{4}\vec{a}+\frac{1}{3}\vec{b}+\frac{1}{6}\vec{c}\)
\(\overrightarrow{OP}=k\overrightarrow{OR}\)より、
\(\displaystyle \overrightarrow{OP}=k\left(\frac{1}{4}\vec{a}+\frac{1}{3}\vec{b}+\frac{1}{6}\vec{c}\right)\)
\(\displaystyle =\frac{1}{4}k\vec{a}+\frac{1}{3}k\vec{b}+\frac{1}{6}k\vec{c}\)
\(A,B,C,P\)は同一平面上にあるので、
\(\displaystyle \frac{1}{4}k+\frac{1}{3}k+\frac{1}{6}k=1\)
\(\displaystyle k=\frac{4}{3}\)
よって、
\(\displaystyle \overrightarrow{OP}=\frac{1}{3}\vec{a}+\frac{4}{9}\vec{b}+\frac{2}{9}\vec{c}\)

\(\displaystyle \overrightarrow{EH}=\overrightarrow{OH}-\overrightarrow{OE}\)
\(\displaystyle =\frac{\overrightarrow{OB}+2\overrightarrow{OC}}{2+1}-\frac{1}{2}\overrightarrow{OA}\)
\(\displaystyle =-\frac{1}{2}\vec{a}+\frac{1}{3}\vec{b}+\frac{2}{3}\vec{c}\)
\(\overrightarrow{EI}=k\overrightarrow{EH}\)より、
\(\displaystyle \overrightarrow{EI}=k\left(-\frac{1}{2}\vec{a}+\frac{1}{3}\vec{b}+\frac{2}{3}\vec{c}\right)\)
\(\displaystyle =-\frac{1}{2}k\vec{a}+\frac{1}{3}k\vec{b}+\frac{2}{3}k\vec{c}\)
\(\displaystyle \overrightarrow{OI}=\overrightarrow{EI}+\overrightarrow{OE}\)
\(\displaystyle =-\frac{1}{2}k\vec{a}+\frac{1}{3}k\vec{b}+\frac{2}{3}k\vec{c}+\frac{1}{2}\vec{a}\)
\(\displaystyle =\frac{1}{2}(1-k)\vec{a}+\frac{1}{3}k\vec{b}+\frac{2}{3}k\vec{c}\)
\(A,F,G,I\)は同一平面上にあるので、
\(\displaystyle \overrightarrow{OI}=\overrightarrow{OA}+\overrightarrow{AI}\)
\(\displaystyle =\vec{a}+s(\overrightarrow{OF}-\overrightarrow{OA})+t(\overrightarrow{OG}-\overrightarrow{OA})\)
\(\displaystyle =\vec{a}+s\left(\frac{1}{2}\vec{b}-\vec{a}\right)+t\left(\frac{1}{2}\vec{c}-\vec{a}\right)\)
\(\displaystyle =(1-s-t)\vec{a}+\frac{1}{2}s\vec{b}+\frac{1}{2}t\vec{c}\)
\(\displaystyle \frac{1}{2}(1-k)=1-s-t,\frac{1}{3}k=\frac{1}{2}s,\frac{2}{3}k=\frac{1}{2}t\)
これを解くと、
\(\displaystyle k=\frac{1}{3},s=\frac{2}{9},t=\frac{4}{9}\)
よって、
\(\displaystyle \overrightarrow{OI}=\frac{1}{3}\vec{a}+\frac{1}{9}\vec{b}+\frac{2}{9}\vec{c}\)