高専の数学解説

第13章 13.7

cos⁡θab=(4,2,2)⋅(4,3−1,−3−1)42+22+2242+(3−1)2+(−3−1)2=16+23−2−23−22424=12\cos\theta_{ab} = \frac{(4,2,2)\cdot(4,\sqrt{3}-1,-\sqrt{3}-1)}{\sqrt{4^2+2^2+2^2}\sqrt{4^2+(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2}} = \frac{16+2\sqrt{3}-2-2\sqrt{3}-2}{\sqrt{24}\sqrt{24}} = \frac{1}{2}

cos⁡θbc=(4,3−1,−3−1)⋅(4,−3−1,3−1)42+(3−1)2+(−3−1)242+(−3−1)2+(3−1)2=122424=12\cos\theta_{bc} = \frac{(4,\sqrt{3}-1,-\sqrt{3}-1)\cdot(4,-\sqrt{3}-1,\sqrt{3}-1)}{\sqrt{4^2+(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2}\sqrt{4^2+(-\sqrt{3}-1)^2+(\sqrt{3}-1)^2}} = \frac{12}{\sqrt{24}\sqrt{24}} = \frac{1}{2}

cos⁡θca=(4,2,2)⋅(4,3−1,−3−1)42+22+2242+(3−1)2+(−3−1)2=16+23−2−23−22424=12\cos\theta_{ca} = \frac{(4,2,2)\cdot(4,\sqrt{3}-1,-\sqrt{3}-1)}{\sqrt{4^2+2^2+2^2}\sqrt{4^2+(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2}} = \frac{16+2\sqrt{3}-2-2\sqrt{3}-2}{\sqrt{24}\sqrt{24}} = \frac{1}{2}

これから θab=θbc=θca=60°\theta_{ab} = \theta_{bc} = \theta_{ca} = 60°

解説: r31bn1z