数学 - 立体的な広がりの中の図形 - 空間図形 – 空間のベクトル(ベクトルの内積、成分表示、なす角、方向余弦)

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数学読本〈3〉平面上のベクトル/複素数と空間のベクトル/空間図形/2次曲線/数列 (松坂和夫(著)、岩波書店)の第11章(立体的な広がりの中の図形 - 空間図形)、11.2(空間のベクトル)、ベクトルの内積、問13、14、15、16、17.を取り組んでみる。

問13

$\begin{array}{l} 1 + 2 = \sqrt{2} \cdot 3 \cos θ \\ \cos θ = \frac{1}{\sqrt{2}} \\ θ = \frac{π}{4} \end{array}$
$\begin{array}{l} 4 + 2 - 6 = \sqrt{14} \cdot \sqrt{21} \cos θ \\ 0 = \cos θ \\ θ = \frac{π}{2} \end{array}$

問14

$\begin{array}{l} \vec{B A} = (1, - 2, 3) \\ \vec{C A} = (- 3, 2, - 1) \\ - 3 - 4 - 3 = \sqrt{14} \sqrt{14} \cos θ \\ \cos θ = - \frac{10}{14} \\ \cos θ = - \frac{5}{7} \end{array}$
$\begin{array}{l} \sin^{2} θ = 1 - \cos^{2} θ = 1 - \frac{25}{49} = \frac{24}{49} \\ \sin θ = \frac{2 \sqrt{6}}{7} \\ \sqrt{14} \sqrt{14} \frac{2 \sqrt{6}}{7} \cdot \frac{1}{2} = 2 \sqrt{6} \end{array}$

問15

\begin{array}{l} S = | \vec{a} | | \vec{b} | \sin θ \\ S^{2} = {| \vec{a} |}^{2} {| \vec{b} |}^{2} \sin^{2} θ \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} (1 - \cos^{2} θ) \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} - {| \vec{a} |}^{2} {| \vec{b} |}^{2} \cos^{2} θ \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} - {(\vec{a} \cdot \vec{b})}^{2} \\ = (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) - {(a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})}^{2} \\ = (b_{2}^{2} + b_{3}^{2}) a_{1}^{2} - 2 (a_{2} b_{2} + a_{3} b_{3}) b_{1} a_{1} + (a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) - {(a_{2} b_{2} + a_{3} b_{3})}^{2} \\ = (b_{2}^{2} + b_{3}^{2}) a_{1}^{2} - 2 (a_{2} b_{2} + a_{3} b_{3}) b_{1} a_{1} + a_{2}^{2} b_{1}^{2} + a_{2}^{2} b_{3}^{2} + a_{3}^{2} b_{1}^{2} + a_{3}^{2} b_{2}^{2} - 2 a_{2} b_{2} a_{3} b_{3} \\ = {(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(a_{2} b_{3} - a_{3} b_{2})}^{2} + {(a_{3} b_{1} - a_{1} b_{3})}^{2} \\ S = \sqrt{(b_{2}^{2} + b_{3}^{2}) a_{1}^{2} - 2 (a_{2} b_{2} + a_{3} b_{3}) b_{1} a_{1} +} \end{array}

問16

\begin{array}{l} \vec{A B} \cdot \vec{C D} \\ = (\vec{b} - \vec{a}) \cdot (\vec{d} - \vec{c}) \\ = \vec{b} \cdot \vec{d} - \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{c} \\ = 0 \end{array}

問17

\begin{array}{l} a_{1} = | \vec{a} | \cos α \\ a_{2} = | \vec{b} | \cos β \\ a_{3} = | \vec{c} | \cos γ \\ \frac{a_{1}}{| \vec{a} |} = \cos α \\ \frac{a_{2}}{| \vec{a} |} = \cos β \\ \frac{a_{3}}{| | \vec{a} | |} = \cos γ \end{array}

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