MathematicsMedium202×since 2002Q5800The vectors a→\overrightarrow aa and b→\overrightarrow bb are not perpendicular and c→\overrightarrow cc and d→\overrightarrow dd are two vectors satisfying b→×c→=b→×d→\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow db×c=b×d and a→.d→=0 .\overrightarrow a .\overrightarrow d = 0\,\,.a.d=0. Then the vector d→\overrightarrow dd is equal to :Ac→+(a→.c→a→.b→)b→\overrightarrow c + \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow bc+(a.ba.c)bBb→+(b→.c→a→.b→)c→\overrightarrow b + \left( {{{\overrightarrow b .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow cb+(a.bb.c)cCc→−(a→.c→a→.b→)b→\overrightarrow c - \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow bc−(a.ba.c)bDb→−(b→.c→a→.b→)c→\overrightarrow b - \left( {{{\overrightarrow b .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow cb−(a.bb.c)cCheck answerSkip