数学 - 線形代数学 - R^nにおけるベクトル – 直線と平面(向き、垂直、平行、距離、交点)

学習環境

ラング線形代数学(上)(S.ラング (著)、芹沢正三 (翻訳)、ちくま学芸文庫)の1章(R^nにおけるベクトル)、5(直線と平面)、練習問題17-20.を取り組んでみる。

$\begin{array}{l} l : (1, 3, - 2) + t (1, 2, 2) \\ x + 2 y + 2 z = d \\ 1 - 2 + 4 = d \\ d = 3 \\ x + 2 y + 2 z = 3 \\ (1 + t) + 2 (3 + 2 t) + 2 (- 2 + 2 t) = 3 \\ 3 + 9 t = 3 \\ t = 0 \\ (1, 3, - 2) \end{array}$
$\begin{array}{l} (1, 3, 5) + t (- 1, 1, - 1) \\ - x + y - z = d \\ 1 + 1 - 7 = d \\ d = - 5 \\ - x + y - z = - 5 \\ - (1 - t) + (3 + t) - (5 - t) = - 5 \\ 3 t = - 2 \\ t = - \frac{2}{3} \\ P' = (1 + \frac{2}{3}, 3 - \frac{2}{3}, 5 + \frac{2}{3}) = (\frac{5}{3}, \frac{7}{3}, \frac{17}{3}) \\ | P P' | = \sqrt{{(\frac{2}{3})}^{2} + {(- \frac{2}{3})}^{2} + {(\frac{2}{3})}^{2}} = \sqrt{\frac{3 \cdot 2^{2}}{3^{2}}} = \frac{2}{\sqrt{3}} \end{array}$
$\begin{array}{l} (1, 3, 5) + t (- 21, 1) \\ 2 (1 - 2 t) + 3 (3 + t) - (5 + t) = 1 \\ 6 - 2 t = 1 \\ t = \frac{5}{2} \\ (1 - 5, 3 + \frac{5}{2}, 5 + \frac{5}{2}) = (- 4, \frac{11}{2}, \frac{15}{2}) \end{array}$
$\begin{array}{l} (1, 1, 2) + t (3, 1, - 5) \\ 3 (1 + 3 t) + (1 + t) - 5 (2 - 5 t) = 2 \\ - 6 + 35 t = 2 \\ t = \frac{8}{35} \\ (1 + \frac{24}{35}, 1 + \frac{8}{35}, 2 - \frac{40}{35}) \\ \sqrt{{(\frac{24}{35})}^{2} + {(\frac{8}{35})}^{2} + {(\frac{40}{35})}^{2}} \\ = \frac{\sqrt{2240}}{35} \\ = \frac{8 \sqrt{35}}{35} \\ = \frac{8}{\sqrt{35}} \end{array}$

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