Question

4£®ÒòΪ|cos£¼$\overrightarrow a$£¬$\overrightarrow b$£¾|¡Ü1£¬ËùÒÔ|$\overrightarrow a$•$\overrightarrow b$|¡Ü|$\overrightarrow a$||$\overrightarrow b$|£¬µ±ÇÒ½öµ±$\overrightarrow a£¬\;\;\overrightarrow b$¹²ÏßʱȡµÈºÅ£¬ÄÇôÈô$\overrightarrow a$=£¨x₁£¬y₁£¬z₁£©£¬$\overrightarrow b$=£¨x₂£¬y₂£¬z₂£©£¬ÔòÓÐ$\sqrt{{{{£¨x}_{1}•x}_{2}£©}^{2}{+{£¨y}_{1}{•y}_{2}£©}^{2}{+{£¨z}_{1}{•z}_{2}£©}^{2}}$¡Ü$\sqrt{{{x}_{1}}^{2}{{+y}_{1}}^{2}{{+z}_{1}}^{2}}$•$\sqrt{{{x}_{2}}^{2}{{+y}_{2}}^{2}{{+z}_{2}}^{2}}$£¬µ±ÇÒ½öµ±µ±$\frac{{x}_{1}}{{x}_{2}}$=$\frac{{y}_{1}}{{y}_{2}}$=$\frac{{z}_{1}}{{z}_{2}}$È¡µÈºÅ£¬ËùÒÔµ±a²+4b²+9c²=6ʱ£¬$\frac{1}{a^2}$+$\frac{1}{b^2}$+$\frac{1}{c^2}$µÄ×îСֵΪ6£®

Answer 1

·ÖÎö ÓÉÌâÒâÀûÓÃÁ½¸öÏòÁ¿µÄÊýÁ¿»ý¹«Ê½ÇóµÃ$\overrightarrow{a}•\overrightarrow{b}$ÒÔ¼°|$\overrightarrow a$||$\overrightarrow b$£¬¿ÉµÃ½áÂÛ£»ÔÙÀûÓûù±¾²»µÈʽÇóµÃÒªÇóʽ×ÓµÄ×îСֵ£¬×¢Òâ1µÄ´ú»»£º$\frac{{a}^{2}}{6}$+$\frac{{2b}^{2}}{3}$+$\frac{{3c}^{2}}{2}$=1£®

½â´ð ½â£º¡ß$\overrightarrow{a}•\overrightarrow{b}$=x₁•x₂ +y₁•y₂ +z₁•z₂ =$\sqrt{{{{£¨x}_{1}•x}_{2}£©}^{2}{+{£¨y}_{1}{•y}_{2}£©}^{2}{+{£¨z}_{1}{•z}_{2}£©}^{2}}$£¬|$\overrightarrow a$||$\overrightarrow b$|=$\sqrt{{{x}_{1}}^{2}{{+y}_{1}}^{2}{{+z}_{1}}^{2}}$•$\sqrt{{{x}_{2}}^{2}{{+y}_{2}}^{2}{{+z}_{2}}^{2}}$£¬
ÓÖ¡ß|$\overrightarrow a$•$\overrightarrow b$|¡Ü|$\overrightarrow a$||$\overrightarrow b$|£¬µ±ÇÒ½öµ±$\overrightarrow a£¬\;\;\overrightarrow b$¹²ÏßʱȡµÈºÅ£¬¡à$\sqrt{{{{£¨x}_{1}•x}_{2}£©}^{2}{+{£¨y}_{1}{•y}_{2}£©}^{2}{+{£¨z}_{1}{•z}_{2}£©}^{2}}$¡Ü$\sqrt{{{x}_{1}}^{2}{{+y}_{1}}^{2}{{+z}_{1}}^{2}}$•$\sqrt{{{x}_{2}}^{2}{{+y}_{2}}^{2}{{+z}_{2}}^{2}}$£¬
 µ±ÇÒ½öµ±$\frac{{x}_{1}}{{x}_{2}}$=$\frac{{y}_{1}}{{y}_{2}}$=$\frac{{z}_{1}}{{z}_{2}}$ Ê±£¬µÈºÅ³ÉÁ¢£®
¡ßa²+4b²+9c²=6£¬¡à$\frac{{a}^{2}}{6}$+$\frac{{2b}^{2}}{3}$+$\frac{{3c}^{2}}{2}$=1£¬¡à$\frac{1}{a^2}$+$\frac{1}{b^2}$+$\frac{1}{c^2}$=£¨ $\frac{1}{6}$+$\frac{{2b}^{2}}{{3a}^{2}}$+$\frac{{3c}^{2}}{{2a}^{2}}$ £©+£¨$\frac{{a}^{2}}{{6b}^{2}}$+$\frac{2}{3}$+$\frac{{3c}^{2}}{{2b}^{2}}$£©+£¨$\frac{{a}^{2}}{{6c}^{2}}$+$\frac{{2b}^{2}}{{3c}^{2}}$+$\frac{3}{2}$£©
=$\frac{7}{3}$+2$\sqrt{\frac{{2b}^{2}}{{3a}^{2}}•\frac{{a}^{2}}{{6b}^{2}}}$+2$\sqrt{\frac{{3c}^{2}}{{2a}^{2}}•\frac{{a}^{2}}{{6c}^{2}}}$+2$\sqrt{\frac{{3c}^{2}}{{2b}^{2}}•\frac{{2b}^{2}}{{3c}^{2}}}$=6£¬µ±ÇÒ½öµ±$\frac{a}{\sqrt{6}}$=$\frac{b}{\sqrt{\frac{3}{2}}}$=$\frac{c}{\sqrt{\frac{2}{3}}}$=$\frac{1}{3}$ʱ£¬È¡µÈºÅ£®
¹Ê´ð°¸Îª£º$\sqrt{{{{£¨x}_{1}•x}_{2}£©}^{2}{+{£¨y}_{1}{•y}_{2}£©}^{2}{+{£¨z}_{1}{•z}_{2}£©}^{2}}$¡Ü$\sqrt{{{x}_{1}}^{2}{{+y}_{1}}^{2}{{+z}_{1}}^{2}}$•$\sqrt{{{x}_{2}}^{2}{{+y}_{2}}^{2}{{+z}_{2}}^{2}}$£»$\frac{{x}_{1}}{{x}_{2}}$=$\frac{{y}_{1}}{{y}_{2}}$=$\frac{{z}_{1}}{{z}_{2}}$£»6£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁ½¸öÏòÁ¿µÄÊýÁ¿»ýµÄÔËË㣬»ù±¾²»µÈʽµÄÓ¦Óã¬ÊôÓÚÖеµÌ⣮