InterviewSolution
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InterviewSolution
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Ab, AC and AD are three adjacent edges of a parallelpiped. The diagonal of the praallelepiped passing through A and direqcted away from it is vector veca. The vector of the faces containing vertices A, B , C and A, B, D are vecb and vecc, respectively , i.e. vec(AB) xx vec(AC) and vec(AD) xx vec(AB) = vecc the projection of each edge AB and AC on diagonal vector veca is |veca|/3 vector vec(AB) is |
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Answer» `1/3 veca+ (vecaxx(VECB-vecc))/|veca|^(2)` `vec(AB)xxvec(AC)=vecb` `vec(AD)xxvec(AB)=vecc` `vec(AB).veca/(|veca|)=|veca|/3 Rightarrowvec(AB).veca= (|veca|^(2))/3` `vec(AB).veca/(|veca|)=|veca|/3 Rightarrowvec(AC).veca= (|veca|^(2))/3` ` (vec(AB) XX vec(AC))xxveca = vecb xxveca` `vec(AC)-vec(AB)=3(vecbxxveca)/(|veca|^(2))` `|veca|^(2)=vec(AB).veca+vec(AC).veca+vec(AD).veca` `(|veca|^(2))/3=vec(AD).veca` `(vec(AD)xxvec(AB))xxveca=veccxxveca` `vec(AB)- vec(AD) = 3 (vecc xx veca)/(|veca|^(2))` Now from (II) and (III), we get `vec(AC) and vec(AD)`as `vec(AC)=1/3veca+ (vecaxx(vecb xx vecc))/(|veca|^(2))+(3(vecbxxveca))/(|veca|^(2))` ` vec(AD)= 1/3veca+ (vecaxx(vecb-vecc))/(|veca|^(2))- (3(vec cxxveca))/(|veca|^(2))` 
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