Question Number 47190 by 23kpratik last updated on 06/Nov/18
$${find}\:{the}\:{angel}\:{between}\:{the}\:{surface}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:\:{and}\:\mathrm{3}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{z}=\mathrm{1}\:{at}\:\left(\mathrm{1},−\mathrm{2},\mathrm{1}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18
$$\overset{\rightarrow} {\bigtriangledown}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$\left({i}\frac{\partial}{\partial{x}}+{j}\frac{\partial}{\partial{y}}+{k}\frac{\partial}{\partial{z}}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$={i}\mathrm{2}{x}+{j}\mathrm{2}{y}+{k}\mathrm{2}{z} \\ $$$$={i}\mathrm{2}\left(\mathrm{1}\right)+{j}\mathrm{2}\left(−\mathrm{2}\right)+{k}\mathrm{2}\left(\mathrm{1}\right) \\ $$$$=\mathrm{2}{i}−\mathrm{4}{j}+\mathrm{2}{k} \\ $$$$\overset{\rightarrow} {\bigtriangledown}\left(\mathrm{3}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}\right) \\ $$$$={i}\left(\mathrm{6}{x}\right)+{j}\left(−\mathrm{2}{y}\right)+{k}\left(\mathrm{2}\right) \\ $$$$={i}\left(\mathrm{6}×\mathrm{1}\right)+{j}\left(−\mathrm{2}×−\mathrm{2}\right)+{k}\left(\mathrm{2}\right) \\ $$$$=\mathrm{6}{i}+\mathrm{4}{j}+\mathrm{2}{k} \\ $$$${cos}\theta=\frac{\mathrm{6}×\mathrm{2}−\mathrm{4}×\mathrm{4}+\mathrm{2}×\mathrm{2}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\:×\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\theta=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$