JEE Main 20207 Jan, Evening shiftMathematicsHardVector Algebra184× since 2002Let a→\overrightarrow aa , b→\overrightarrow bb and c→\overrightarrow cc be three unit vectors such that a→+b⃗+c→=0→\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0a+b+c=0. If λ=a→.b⃗+b⃗.c→+c→.a→\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow aλ=a.b+b.c+c.a and d→=a→×b⃗+b⃗×c→+c→×a→\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow ad=a×b+b×c+c×a, then the ordered pair, (λ,d→)\left( {\lambda ,\overrightarrow d } \right)(λ,d) is equal to :AOption A. (32,3a→×c→)\left( {{3 \over 2},3\overrightarrow a \times \overrightarrow c } \right)(23,3a×c)BOption B. (−32,3c→×b→)\left( { - {3 \over 2},3\overrightarrow c \times \overrightarrow b } \right)(−23,3c×b)COption C. (−32,3a→×b→)\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)(−23,3a×b)DOption D. (32,3b→×c→)\left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right)(23,3b×c)CheckSkip · show solution
