<prompt> Acha mathbfa, mathbfb, mathbfc, mathbfd ziwe vivekta vinne tofauti katika nafasi ilimathbfa cdot mathbfb mathbfa cdot mathbfc mathbfb cdot mathbfc mathbfb cdot mathbf d mathbfc cdot mathbfd -frac111.Tafuta mathbfa cdot mathbfd.  <response>: Acha O iwe asili, na A,B,C,D iwe pointi kwenye nafasi ili overrightarrowOA mathbfa, overrightarrowOB mathbfb, overrightarrowOC mathbfc, na overrightarrowOD mathbfd.asykuagiza tatu;ukubwa 180;makadirio ya sasa mtazamo 6,3,2;mara tatu A, B, C, D, O;A -1sqrt55,0,3sqrt6sqrt55;B sqrt511, -sqrt611, 0;C sqrt511, sqrt611, 0;D -1sqrt55,0,-3sqrt6sqrt55;O 0,0,0;choraO--A,Arrow36;choraO--B,Arrow36;choraO--C,Arrow36;choraO--D,Arrow36;choraA--B--D--C--mzunguko, umekatika;kuchoraB--C,dashed;leboA, A, N;leboB, B, W;leboC, C, SE;leboD, D, S;leboO, O, NW;lebomathbfa, A2, W;lebomathbfb, B2, N;lebomathbfc, C2, NE;lebomathbfd, D2, W;asyKumbuka kuwa cos angle AOB -frac111, hivyo kwa Sheria ya Cosines kwenye pembetatu AOB,AB sqrt1 1 - 211 kushoto -frac111 kulia sqrtfrac2411 2 sqrtfrac611.Vile vile, AC BC BD CD 2 sqrtfrac611.Acha M iwe katikati ya overlineBC. Kwa kuwa pembetatu ABC ni sawa na urefu wa upande 2 sqrtfrac611, BM CM sqrt frac611, na AM sqrt3 cdot sqrtfrac611 sqrtfrac1811.asykuagiza tatu;ukubwa 180;makadirio ya sasa mtazamo 6,3,2;mara tatu A, B, C, D, M, O;A -1sqrt55,0,3sqrt6sqrt55;B sqrt511, -sqrt611, 0;C sqrt511, sqrt611, 0;D -1sqrt55,0,-3sqrt6sqrt55;O 0,0,0;M B C2;kuchoraO--A,dashed;kuchoraO--B,dashed;kuchoraO--C,dashed;kuchoraO--D,dashed;kuchoraA--B--D--C--mzunguko;kuchoraB--C;kuchoraA--M;kuchoraM--O,dashed;leboA, A, N;leboB, B, W;leboC, C, SE;leboD, D, S;leboM, M, S;leboO, O, NW;asyKisha na Pythagoras kwenye pembetatu ya kulia BMO,MO sqrtBO2 - BM2 sqrt1 - frac611 sqrtfrac511.Kwa Sheria ya Cosines kwenye pembetatu AMO,cos angle AOM fracAO2 MO2 - AM22 cdot AO cdot MO frac1 frac511 - frac 18112 cdot 1 cdot sqrtfrac511 -frac1sqrt55.Kishaanzaalignmathbfa cdot mathbfd cos angle AOD cos 2 pembe AOM 2 cos2 pembe AOM - 1 2 kushoto -frac1sqrt55 kulia2 - 1 -frac5355.mwishopanga 