こんにちは。今回はベクトルと内分点ということで, 比較的重要な例題をやっておきます。それでは見ていきましょう。
【例】△OABにおいて, 辺OAをに内分する点をC, 線分BCを
に内分する点をDとし, 直線ODと辺ABの交点をEとする。このとき,
を
,
で表し, OD : DEを求めよ。
【解法】OD : DEだけならメネラウスの定理を使えば出せますが, それは今回は無しということでいきます。
CD : DB=1 : 2なので,
![Rendered by QuickLaTeX.com \overrightarrow{\text{OD}}=\dfrac23\overrightarrow{\text{OC}}+\dfrac13\overrightarrow{\text{OB}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-194de1c21a715ede615d1f0b78aad433_l3.png)
![Rendered by QuickLaTeX.com \overrightarrow{\text{OC}}=\dfrac35\overrightarrow{\text{OA}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-701906ac1b8a57ecfa901c9dc7a1c810_l3.png)
![Rendered by QuickLaTeX.com \begin{array}{lll}\overrightarrow{\text{OD}}&=&\dfrac23\cdot\dfrac35\overrightarrow{\text{OA}}+\dfrac13\overrightarrow{\text{OB}}\\&=&\dfrac15\overrightarrow{\text{OA}}+\dfrac23\overrightarrow{\text{OB}}\end{array}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-58f90e6e09d5250ad08e1648379fdc38_l3.png)
よって,
![Rendered by QuickLaTeX.com \overrightarrow{\text{OD}}=\dfrac25\overrightarrow{\text{OA}}+\dfrac13\overrightarrow{\text{OB}}\cdots](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-633905c264f9a4cd5768d6dac506525f_l3.png)
ここで,
![Rendered by QuickLaTeX.com \overrightarrow{\text{OE}}=k\overrightarrow{\text{OD}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-feb4c23868b7d11fe8196943fdaab792_l3.png)
![Rendered by QuickLaTeX.com \begin{array}{lll}\overrightarrow{\text{OE}}&=&k\left(\dfrac25\overrightarrow{\text{OA}}+\dfrac13\overrightarrow{\text{OB}}\right)\\&=&\dfrac25k\overrightarrow{\text{OA}}+\dfrac13k\overrightarrow{\text{OB}}\end{array}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-365a1cb250aa72b7ad37b90a710f6d96_l3.png)
Eは辺AB上にあるので,
![Rendered by QuickLaTeX.com \dfrac25k+\dfrac13k=1](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-56f283e63b149a31685cd4242fd6a240_l3.png)
これを解いて,
![Rendered by QuickLaTeX.com k=\dfrac{15}{11}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-04b035ef9b89fb161db28497a910a423_l3.png)
よって,
![Rendered by QuickLaTeX.com \overrightarrow{\text{OE}}=\dfrac{15}{11}\overrightarrow{\text{OD}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-59aa18861d70829dbd84dfbeba1058b0_l3.png)
したがって, OD : DE
![Rendered by QuickLaTeX.com =11 : 4\cdots](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-2e5f2e3b4188a9f24f356e95a626dc01_l3.png)
【別解】
![Rendered by QuickLaTeX.com \overrightarrow{\text{OD}}=\dfrac25\overrightarrow{\text{OA}}+\dfrac13\overrightarrow{\text{OB}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-bad3b8e69dc54b0b7f33c25ca4faa98b_l3.png)
を次のように変形することもできる。
![Rendered by QuickLaTeX.com \begin{array}{lll}\overrightarrow{\text{OD}}&=&\dfrac25\overrightarrow{\text{OA}}+\dfrac13\overrightarrow{\text{OB}}\\&=&\dfrac{6}{15}\overrightarrow{\text{OA}}+\dfrac{5}{15}\overrightarrow{\text{OB}}\\&=&\dfrac{6+5}{15}\left(\dfrac{6\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}}{11}\right)\\&=&\dfrac{11}{15}\left(\underline{\dfrac{6\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}}{11}}\right)\end{array}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-d3a80afe78fa74cdbc390f9404f878b6_l3.png)
下線部は
![Rendered by QuickLaTeX.com \overrightarrow{\text{OE}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-21687969476834466e1bb5a9321cbf1b_l3.png)
![Rendered by QuickLaTeX.com \overrightarrow{\text{OD}}=\dfrac{11}{15}\overrightarrow{\text{OE}}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-ac2aac5d17b8a47c7fe16e95c545a361_l3.png)
よって, OD : DE
![Rendered by QuickLaTeX.com =11 : 4\cdots](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-2e5f2e3b4188a9f24f356e95a626dc01_l3.png)