Question

Let \(\left[ {\vec a} \right] = {\rm{\hat i}} + {\rm{\hat j}} + \sqrt 2 \widehat {{\rm{k\;}}},\;\vec b = {{\rm{b}}_1}{\rm{\hat i}} + {{\rm{b}}_2}{\rm{\hat j}} + \sqrt 2 \widehat {{\rm{k\;}}}{\rm{\;and\;}}\vec c = 5{\rm{\hat i}} + {\rm{\hat j}} + \sqrt 2 \widehat {{\rm{k}}}\) be three vectors such that the projection vector of \(\vec b{\rm{\;on\;}}\vec a{\rm{\;is\;}}\vec a{\rm{.\;If\;}}\vec a + \vec b\)is perpendicular to \(\vec c,{\rm{\;then\;}}\left| {\vec b} \right|\) is equal to:
1. \(\sqrt {32}\)
2. 6
3. \(\sqrt {22}\)
4. 4

Answer 1

Correct Answer - Option 2 : 6

Projection of \(\vec b{\rm{\;on\;}}\vec a = \frac{{\vec b\cdot\vec a}}{{\left| {\vec a} \right|}} = \frac{{{b_1} + {b_2} + 2}}{4}\) 

According to the question,

\(\frac{{{b_1} + {b_2} + 2}}{2} = \sqrt {1 + 1 + 2} = 2\) 

⇒ b₁ + b₂ = 2     ....(1)

Since, \(\vec a + \vec b{\rm{\;is\;perpendicular\;to\;}}\vec c\).

Hence, \(\vec a.\vec c + \vec b.\vec c = 0\)

⇒ 8 + 5b₁ + b₂ + 2 = 0     ....(2)

From (1) and (2)

b₁ = – 3, b₂ = 5

\(\Rightarrow \vec b = - 3\hat i + 5{\rm{\hat j}} + \sqrt 2 \widehat {{\rm{k}}}\) 

\(\Rightarrow \left| {\vec b} \right| = \sqrt {{3^2} + {5^2} + {{\left( {\sqrt 2 } \right)}^2}}\) 

\(\left| {\vec b} \right| = \sqrt {9 + 25 + 2} = 6\)