MCQ
If $\vec{a},\vec{b},\vec{c}$are three non-zero, non-coplanar vectrors and $\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 \, $ and $ \overrightarrow {{c_1}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow a \,} \right|}^2}}}\overrightarrow a \, + \,\frac{{\overrightarrow c \,.\,\overrightarrow b }}{{{{\left| {\overrightarrow b \,} \right|}^2}}}\overrightarrow {{b_1}} \, $, $\overrightarrow {{c_2}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow a \,} \right|}^2}}}\overrightarrow a \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow b }}{{{{\left| {\overrightarrow {{b_1}} \,} \right|}^2}}}\overrightarrow {{b_1}} \, ,$ $ \overrightarrow {{c_3}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow c \,} \right|}^2}}}\overrightarrow a \, + \,\frac{{\overrightarrow c \,.\,\overrightarrow {{b_2}} }}{{{{\left| {\overrightarrow c \,} \right|}^2}}}\overrightarrow {{b_1}} \, $ $, \overrightarrow {{c_4}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow c \,} \right|}^2}}}\overrightarrow a \, - \,\frac{{\overrightarrow b \,.\,\overrightarrow c }}{{{{\left| {\overrightarrow b \,} \right|}^2}}}\overrightarrow {{b_1}} \,.$ Then, which of the following is a set of mutually orthogonal vectors ?
  • A
    {$\overrightarrow {a,} \overrightarrow {{b_1},} \overrightarrow {{c_1}}$} 
  • {$\overrightarrow {a,} \overrightarrow {{b_1},} \overrightarrow {{c_2}}$} 
  • C
    {$\overrightarrow {a,} \overrightarrow {{b_2},} \overrightarrow {{c_3}}$} 
  • D
    {$\overrightarrow {a,} \overrightarrow {{b_2},} \overrightarrow {{c_4}}$}

Answer

Correct option: B.
{$\overrightarrow {a,} \overrightarrow {{b_1},} \overrightarrow {{c_2}}$} 
b
$\overrightarrow {\rm{a}}  \cdot {\overrightarrow {\rm{b}} _1} = \vec a \cdot \left( {\vec b - \frac{{\vec b \cdot \vec a}}{{|\vec a{|^2}}}\vec a} \right)$

$ = \vec a \cdot \vec b - \frac{{|\vec a{|^2}(\vec b \cdot \vec a)}}{{{{\left| {\vec a} \right|}^2}}}$

$ = \vec a \cdot \vec b - \vec b \cdot \vec a = 0$

Similarly, $\vec a \cdot {{\vec c}_2} = {{\vec b}_1} \cdot {{\vec c}_2} = 0$

Hence, $\left\{ {\vec a,{{\vec b}_1},{{\vec c}_2}} \right\}$ are mutually orthogonal vectors.

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