数学 - 線型代数 - 2次元と3次元の簡単な幾何学 - 空間における直線・平面の方程式(直交する空間の2つの単位ベクトル、任意のベクトル、距離、最小値)

学習環境

線型代数入門(松坂和夫(著)、岩波書店)の第1章(2次元と3次元の簡単な幾何学)、10(空間における直線・平面の方程式)、問15.を取り組んでみる。

1. $\begin{array}{l} (a - c_{1} u_{1} - c_{2} u_{2}) \cdot u_{1} \\ = a \cdot u_{1} - c_{1} u_{1} \cdot u_{1} - c_{2} u_{2} \cdot u_{1} \\ = c_{1} - c_{1} - c_{2} 0 \\ = 0 \\ (a - c_{1} u_{1} - c_{2} u_{2}) \cdot u_{2} \\ = a \cdot u_{2} - c_{1} u_{1} \cdot u_{2} - c_{2} u_{2} \cdot u_{2} \\ = c_{2} - c_{1} 0 - c_{2} \\ = 0 \end{array}$
2. $\begin{array}{l} {| a - t_{1} u_{1} - t_{2} u_{2} |}^{2} \\ = (a - t_{1} u_{1} - t_{2} u_{2}) \cdot (a - t_{1} u_{1} - t_{2} u_{2}) \\ = {| a |}^{2} - t_{1} c_{1} - t_{2} c_{2} - t_{1} c_{1} + t_{1}^{2} - t_{2} c_{2} + t_{2}^{2} \\ = {| a |}^{2} - 2 t_{1} c_{1} - 2 t_{2} c_{2} + t_{1}^{2} + t_{2}^{2} \\ = {| a |}^{2} + {(t_{1} - c_{1})}^{2} + {(t_{2} - c_{2})}^{2} - c_{1}^{2} - c_{2}^{2} \\ = {(t_{1} - c_{1})}^{2} + {(t_{2} - c_{2})}^{2} + ({| a |}^{2} - c_{1}^{2} - c_{2}^{2}) \\ = {(t_{1} - c_{1})}^{2} + {(t_{2} - c_{2})}^{2} + ({| a |}^{2} - 2 c_{1}^{2} - 2 c_{2}^{2} + c_{1}^{2} + c_{2}^{2}) \\ = {(t_{1} - c_{1})}^{2} + {(t_{2} - c_{2})}^{2} + {| a - c_{1} u_{1} - c_{1} u_{2} |}^{2} \end{array}$
  よって最小になる場合。
  $t_{1} = c_{1}, t_{2} = c_{2}$

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