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Question Number 108947 by mohammad17 last updated on 20/Aug/20

Answered by 1549442205PVT last updated on 21/Aug/20

$$\mathrm{We}\mathrm{find}\mathrm{the}\mathrm{angle}\alpha ,\beta ,\gamma \mathrm{between}\mathrm{the}\mathrm{vector}$$$$\overrightarrow{\mathrm{m}}=(1,2,1)\mathrm{and}\mathrm{the}\mathrm{axises}\mathrm{Ox},\mathrm{Oy},\mathrm{Oz}$$$$\text{Missing left or extra right}$$$$\frac{\mid 1.1+2.0+1.0\mid }{\sqrt{1^2+2^2+1^2.1}}=\frac{1}{\sqrt{6}}\Rightarrow \sin \alpha =\sqrt{\frac{5}{6}}$$$$\cos \beta =<\overrightarrow{\mathrm{m}}\overrightarrow{.\mathrm{j}}>/\sqrt{6}=2/\sqrt{6},\cos \gamma =1/\sqrt{6}$$$$\frac{\mathrm{df}}{\overrightarrow{\mathrm{dm}}}=\frac{\mathrm{df}}{\mathrm{dx}}\cos \alpha +\frac{\mathrm{df}}{\mathrm{dy}}\cos \beta +\frac{\mathrm{df}}{\mathrm{dz}}\cos \gamma =$$$$=(\frac{2\mathrm{xy}}{\sqrt{6}}+\frac{({\mathrm{x}}^2-2\mathrm{z})}{\sqrt{6}}+\frac{2\mathrm{z}-2\mathrm{y}}{\sqrt{6}}){\mid }_{(2,1,3)}$$$$=\frac{4}{\sqrt{6}}+\frac{-4}{\sqrt{6}}+\frac{4}{\sqrt{6}}=1.632\Rightarrow \mathrm{Choose}\mathbf{C}$$

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