←Vrati se na zadatke1. ZadatakAko vrijede jednakosti xa+yb+zc=0iax+by+cz=1,\frac xa+\frac yb+\frac zc=0\quad\text{i}\quad\frac ax+\frac by+\frac cz=1,ax+by+cz=0ixa+yb+zc=1, dokaži da vrijedi a2x2+b2y2+c2z2=1.\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=1.x2a2+y2b2+z2c2=1.Prikaži rješenjeRješenjeKvadriranjem druge jednakosti dobivamo 1=(ax+by+cz)2=a2x2+b2y2+c2z2+2(bayz+cazx+abxy)=a2x2+b2y2+c2z2+(bayz+cazx+abxy)=a2x2+b2y2+c2z2+2⋅abcxyz(xa+yb+zc)=a2x2+b2y2+c2z2.\begin{aligned} 1&=\left(\frac ax+\frac by+\frac cz\right)^2\\ &=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\left(\frac{ba}{yz}+\frac{ca}{zx}+\frac{ab}{xy}\right)\\ &=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+\left(\frac{ba}{yz}+\frac{ca}{zx}+\frac{ab}{xy}\right)\\ &=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\cdot\frac{abc}{xyz}\left(\frac xa+\frac yb+\frac zc\right)\\ &=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}. \end{aligned}1=(xa+yb+zc)2=x2a2+y2b2+z2c2+2(yzba+zxca+xyab)=x2a2+y2b2+z2c2+(yzba+zxca+xyab)=x2a2+y2b2+z2c2+2⋅xyzabc(ax+by+cz)=x2a2+y2b2+z2c2.Pomakni formulu lijevo ili desno.