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Vector Algebra question

2011 · Shift 0 · Q35
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  5. /2011 · Shift 0 · Q35

Vector Algebra question

2011 · Shift 0 · Q35

JEE MainMathematicsVector AlgebraMCQ+4 / −1
The 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 :
  1. A
    c→+(a→.c→a→.b→)b→\overrightarrow c + \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow bc+(a.ba.c​)b
  2. B
    b→+(b→.c→a→.b→)c→\overrightarrow b + \left( {{{\overrightarrow b .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow cb+(a.bb.c​)c
  3. C
    c→−(a→.c→a→.b→)b→\overrightarrow c - \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow bc−(a.ba.c​)b
  4. D
    b→−(b→.c→a→.b→)c→\overrightarrow b - \left( {{{\overrightarrow b .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow cb−(a.bb.c​)c
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