高専の数学解説

第13章 13.39

(1)

BA→=(3−2,0+1,−3−1)=(1,1,−4)BC→=(1−2,1+1,−1−1)=(−1,2,−2)\overrightarrow{BA} = (3-2,0+1,-3-1) = (1,1,-4) \qquad \overrightarrow{BC} = (1-2,1+1,-1-1) = (-1,2,-2)

cos⁡(∠B)=BA→⋅BC→∣BA→∣∣BC→∣=1⋅(−1)+1⋅2+(−4)⋅(−2)12+12+(−4)2(−1)2+22+(−2)2=918⋅3=12\cos(\angle B) = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{\left|\overrightarrow{BA}\right|\left|\overrightarrow{BC}\right|} = \frac{1\cdot(-1)+1\cdot2+(-4)\cdot(-2)}{\sqrt{1^2+1^2+(-4)^2}\sqrt{(-1)^2+2^2+(-2)^2}} = \frac{9}{\sqrt{18}\cdot 3} = \frac{1}{\sqrt{2}}

∠B=45°\angle B = 45°

(2)

BA→=(1+1,0−3,1−0)=(2,−3,1)BC→=(2+1,2−3,−2−0)=(3,−1,−2)\overrightarrow{BA} = (1+1,0-3,1-0) = (2,-3,1) \qquad \overrightarrow{BC} = (2+1,2-3,-2-0) = (3,-1,-2)

cos⁡(∠B)=BA→⋅BC→∣BA→∣∣BC→∣=2⋅3+(−3)⋅(−1)+1⋅(−2)22+(−3)2+1232+(−1)2+(−2)2=71414=12\cos(\angle B) = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{\left|\overrightarrow{BA}\right|\left|\overrightarrow{BC}\right|} = \frac{2\cdot3+(-3)\cdot(-1)+1\cdot(-2)}{\sqrt{2^2+(-3)^2+1^2}\sqrt{3^2+(-1)^2+(-2)^2}} = \frac{7}{\sqrt{14}\sqrt{14}} = \frac{1}{2}

∠B=60°\angle B = 60°

解説: r31bn1z