https://mathmanstudio.com/categories/%E5%B9%B3%E9%9D%A2%E5%90%91%E9%87%8F/Recent content in 平面向量 on MathManStudio.comHugozh-TWMon, 13 Jul 2026 20:35:58 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B811/Mon, 13 Jul 2026 20:35:58 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B811/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-13-%E6%99%9A%E4%B8%8A7.46.06-1024x397.png"></p> <p>（1）根據內分比性質可得</p> <p>$$ \overline{CP}:\overline{PD}=4:3 $$ $$ \Rightarrow\overline{CP}=\frac{4}{7}\overline{CD} $$</p> <p>(2)利用向量加法性質轉換，做出線性組合。</p> <p>$$ \overrightarrow{AP}＝\overrightarrow{AC}+\overrightarrow{CP}=\overrightarrow{AC}+\frac{4}{7}\overrightarrow{CD}=\overrightarrow{AC}+\frac{4}{7}(\overrightarrow{CA}+\overrightarrow{AD})=\frac{3}{7}\overrightarrow{AC}+\frac{2}{7}\overrightarrow{AB} $$</p> <p>(3)餘弦定理，送分啦！</p> <p>$$ cos\angle{BAC}=\frac{25+36-16}{2\times5\times6}=\frac{3}{4} $$</p> <p>(4)三邊長都給你了，還有別的嗎？當然是海龍公式，送分啦！</p> <p>$$ \triangle{ABC}_a=\sqrt{\frac{15}{2}\times\frac{5}{2}\times\frac{3}{2}\times\frac{7}{2}}=\frac{15}{4}\sqrt{7} $$ $$ \triangle{ACP}_a=\frac{15}{4}\sqrt{7}\times\frac{1}{2}\times\frac{4}{7}=\frac{15}{14}\sqrt{7} $$</p> <p>(5)高中數學的多選題後面的選項，不要空想，常常會用到前面算出來的東西，前面做不出來後面就很難拿分！</p> <p>$$ \overrightarrow{AP}\cdot\overrightarrow{AC}=(\frac{3}{7}\overrightarrow{AC}+\frac{2}{7}\overrightarrow{AB})\cdot\overrightarrow{AC}=\frac{120}{7} $$</p> <p>Ans:(3)(4)(5)</p>https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a-%E5%A4%9A%E9%81%B88/Wed, 01 Jul 2026 20:36:41 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a-%E5%A4%9A%E9%81%B88/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-01-%E6%99%9A%E4%B8%8A7.33.00-1024x397.png"></p> <p>將原式配方整理後可得圓方程式如下：</p> <p>$$ （x-1）^2+y^2=4 $$</p> <p>此為圓心（1,0）半徑=2之圓方程式</p> <p>$$ (1)直線方程式x=3與此圓方程式之切線，交於1點，有兩相等實根。 $$</p> <p>（2）此圖形並未於原點有點對稱的現象。</p> <p>（3）正確。</p> <p>（4）解方程式判斷有無交點即可。</p> <p>$$ y=-x+4\quad代入（x-1）^2+y^2=4 $$ $$ 整理後可得x^2-5x+\frac{13}{2}=0 $$ $$ 判別式&lt;0\rightarrow無交點 $$</p> <p>(5)求極值當然用柯西不等式啦！</p> <p>$$ 令\vec{a}=(x-1,y)\quad\vec{b}=(1,-1) $$ $$ [(x-1)^2+y^2]\times[1^2+(-1)^2]\ge（x-1-y）^2 $$ $$ 1-2\sqrt{2}\le x-y\le1+2\sqrt{2} $$</p> <p>正解：（3）（5）</p>https://mathmanstudio.com/posts/113%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%88%86%E7%A7%91%E6%B8%AC%E9%A9%97%E7%B7%B4%E5%8A%9F%E5%9D%8A-%E4%BA%8C%E6%AC%A1%E6%9B%B2%E7%B7%9A%E5%B9%B3%E9%9D%A2%E5%90%91%E9%87%8F%E5%80%92%E6%95%B824/Sat, 15 Jun 2024 00:30:21 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