Question
If $$\vec a,\,\vec b,\,\vec c$$ are three non-zero, non-coplanar vectors and
$$\eqalign{
& \overrightarrow {{b_1}} = \overrightarrow b - \frac{{\overrightarrow b .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a ,\,\,\overrightarrow {{b_2}} = \overrightarrow b + \frac{{\overrightarrow b .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a , \cr
& \overrightarrow {{c_1}} = \overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a + \frac{{\overrightarrow b .\overrightarrow c }}{{{{\left| {\overrightarrow c } \right|}^2}}}\overrightarrow {{b_1}} ,\,\,\overrightarrow {{c_2}} = \overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a - \frac{{\overrightarrow {{b_1}} .\overrightarrow c }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} , \cr
& \overrightarrow {{c_3}} = \overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow c } \right|}^2}}}\overrightarrow a + \frac{{\overrightarrow b .\overrightarrow c }}{{{{\left| {\overrightarrow c } \right|}^2}}}\overrightarrow {{b_1}} ,\,\,\overrightarrow {{c_4}} = \overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow c } \right|}^2}}}\overrightarrow a = \frac{{\overrightarrow b .\overrightarrow c }}{{{{\left| {\vec b} \right|}^2}}}\overrightarrow {{b_1}} , \cr} $$
then the set of orthogonal vectors is :
A.
$$\left( {\overrightarrow a ,\,\overrightarrow {{b_1}} ,\,\overrightarrow {{c_3}} } \right)$$
B.
$$\left( {\overrightarrow a ,\,\overrightarrow {{b_1}} ,\,\overrightarrow {{c_2}} } \right)$$
C.
$$\left( {\overrightarrow a ,\,\overrightarrow {{b_1}} ,\,\overrightarrow {{c_1}} } \right)$$
D.
$$\left( {\overrightarrow a ,\,\overrightarrow {{b_2}} ,\,\overrightarrow {{c_2}} } \right)$$
Answer :
$$\left( {\overrightarrow a ,\,\overrightarrow {{b_1}} ,\,\overrightarrow {{c_2}} } \right)$$
Solution :
We observe that
$$\eqalign{
& \overrightarrow a .\overrightarrow {{b_1}} = \overrightarrow a .\overrightarrow b - \left( {\frac{{\overrightarrow {b.} \overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}} \right)\overrightarrow a .\overrightarrow a = \overrightarrow a .\overrightarrow b - \overrightarrow a .\overrightarrow b = 0 \cr
& \overrightarrow a .\overrightarrow {{c_2}} = \overrightarrow a \left( {\overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} } \right) \cr
& = \overrightarrow a .\overrightarrow c - \overrightarrow c .\frac{{\overrightarrow a .\overrightarrow c }}{{{{\left| {\overrightarrow a } \right|}^2}}}{\left| {\overrightarrow a } \right|^2} - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\left( {\overrightarrow a .\overrightarrow {{b_1}} } \right) \cr
& = \overrightarrow a .\overrightarrow c - \overrightarrow a .\overrightarrow c - 0\,\,\,\,\,\,\,\,\,\left[ {\because \,\,\overrightarrow a .\overrightarrow {{b_1}} = 0} \right] \cr
& = 0 \cr
& {\text{And }}\overrightarrow {{b_1}} .\overrightarrow {{c_2}} = \overrightarrow {{b_1}} .\left( {\overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} } \right) \cr
& = \overrightarrow {{b_1}} .\overrightarrow c - \frac{{\left( {\overrightarrow c .\overrightarrow a } \right)\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)}}{{{{\left| {\overrightarrow a } \right|}^2}}} - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} .\overrightarrow {{b_1}} \cr
& = \overrightarrow {{b_1}} .\overrightarrow c - 0 - \overrightarrow {{b_1}} .\overrightarrow c \,\,\,\,\,\,\,\,\,\left[ {{\text{Using }}\overrightarrow {{b_1}} .\overrightarrow a = 0} \right] \cr
& = 0 \cr} $$
Hence $$\overrightarrow a .\overrightarrow {{b_1}} = \overrightarrow a .\overrightarrow {{c_2}} = \overrightarrow {{b_1}} .\overrightarrow {{c_2}} = 0$$
$$ \Rightarrow \left( {\overrightarrow a ,\overrightarrow {{b_1}} ,\,\overrightarrow {{c_2}} } \right)$$ is a set of orthogonal vectors.