平面に関連した問題

■問題

${\displaystyle 3}$ 点${\displaystyle \text{A}\left(3,1,0\right)}$ ，${\displaystyle \text{B}\left(2,0,1\right)}$ ，${\displaystyle \text{C}\left(0,1,1\right)}$ がある．以下の問に答えよ．

(1)${\displaystyle \overrightarrow{\text{AB}}}$ ，${\displaystyle \overrightarrow{\text{AC}}}$ を求めよ．

(2) ${\displaystyle \angle \text{BAC}=\theta }$ とする． ${\displaystyle \cos \theta }$ の値を求めよ．

(3) ${\displaystyle \overrightarrow{\text{AB}}}$ に平行で大きさが ${\displaystyle 1}$ のベクトル ${\displaystyle a}$ を求めよ．

(4)${\displaystyle 2}$ 点${\displaystyle \text{A}}$ ，${\displaystyle \text{B}}$ を通る直線の方程式を求めよ．

(5) ${\displaystyle \overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}}}$ を求めよ．

(6) ${\displaystyle 3}$ 点 ${\displaystyle \text{A}}$ ， ${\displaystyle \text{B}}$ ， ${\displaystyle \text{C}}$ を通る平面の方程式を求めよ．

(7) ${\displaystyle △\text{ABC}}$ の面積を求めよ．

■答

(1)

${\displaystyle \overrightarrow{\text{AB}}=\left(2,0,1\right)-\left(3,1,0\right)=\left(-1,-1,1\right)}$

${\displaystyle \overrightarrow{\text{AC}}=\left(0,1,1\right)-\left(3,1,0\right)=\left(-3,0,1\right)}$

(2)

${\displaystyle \cos \theta =\frac{\overrightarrow{\text{AB}}·\overrightarrow{\text{AC}}}{\left|\overrightarrow{\text{AB}}\right|\left|\overrightarrow{\text{AC}}\right|}}$ ${\displaystyle =\frac{-1\times \left(-3\right)+\left(-1\right)\times 0+1\times 1}{\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2+1^2}·\sqrt{{\left(-3\right)}^2+1^2}}}$ ${\displaystyle =\frac{3+1}{\sqrt{3}·\sqrt{10}}=\frac{4}{\sqrt{30}}}$ ${\displaystyle =\frac{4\sqrt{30}}{30}=\frac{2\sqrt{30}}{15}}$

(3)

${\displaystyle \left|\overrightarrow{\text{AB}}\right|=\sqrt{3}}$

${\displaystyle a=\pm \frac{1}{\sqrt{3}}\left(-1,-1,1\right)}$ ${\displaystyle =\pm \left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)}$ ${\displaystyle =\pm \left(-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\right)}$

(4)

あるいは

(5)

${\displaystyle \overrightarrow{\text{AB}}=\left(-1,-1,1\right)}$

${\displaystyle \overrightarrow{\text{AC}}=\left(-3,0,1\right)}$

(6)

${\displaystyle -1\left(x-3\right)-2\left(y-1\right)-3z}$${\displaystyle =0}$

${\displaystyle -x+3-2y+2-3z}$${\displaystyle =0}$

${\displaystyle -x-2y-3z+5}$${\displaystyle =0}$

${\displaystyle x+2y+3z-5}$${\displaystyle =0}$

■別解

${\displaystyle -1\left(x-2\right)-2y-3\left(z-1\right)}$${\displaystyle =0}$

${\displaystyle -x+2-2y-3z+3}$${\displaystyle =0}$

${\displaystyle -x-2y-3z+5}$${\displaystyle =0}$

${\displaystyle x+2y+3z-5}$${\displaystyle =0}$

(7)

${\displaystyle △\text{ABC}}$ の面積は${\displaystyle \frac{1}{2}\left|\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}}\right|}$ である．

よって

${\displaystyle \frac{1}{2}\sqrt{{\left(-1\right)}^2+{\left(-2\right)}^2+{\left(-3\right)}^2}}$ ${\displaystyle =\frac{1}{2}\sqrt{1+4+9}}$ ${\displaystyle =\frac{\sqrt{14}}{2}}$

　

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最終更新日： 2023年2月16日