MCQ
If $a,b,c$  are three non-zero, non-coplanar vectors and

${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$ , ${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$

, ${{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}$

,${{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}$,

${{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a$.

Then which of the following is a set of mutually orthogonal vectors is

  • A
    $\{{a},\,{b_{1}},\,{{c}_{1}}\}$
  • $\{{a},\,{b_{1}},\,{{c}_{2}}\}$
  • C
    $\{{a},\,{b_{2}},\,{{c}_{3}}\}$
  • D
    $\{{a},\,{b_{2}},\,{{c}_{4}}\}$

Answer

Correct option: B.
$\{{a},\,{b_{1}},\,{{c}_{2}}\}$
b
(b) We have, $a\,.\,{b_1} = 0$, ${b_1}\,.\,{c_2} = 0$, $a\,.\,{c_2} = 0$

$\therefore $ Set of orthogonal vectors, $[a\,\,{b_1}\,\,{c_2}] = 0$

$\therefore $ Option $(b)$  is the correct answer.

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