A) 4
B) \[\sqrt{22}\]
C) 6
D) \[\sqrt{32}\]
Correct Answer: C
Solution :
Projection or \[\overrightarrow{b}\] on \[\overrightarrow{a}\] is \[\overrightarrow{a}\] \[\therefore \,\,\,\frac{\overrightarrow{b}\,.\,\overrightarrow{a}}{\left| \overrightarrow{a} \right|}\,\,=\,\,\left| \overrightarrow{a} \right|\] \[\Rightarrow \frac{{{b}_{1}}+{{b}_{2}}+2}{2}=2\] \[{{b}_{1}}+{{b}_{2}}=2\] .... (1) and \[\overrightarrow{a}+\overrightarrow{b}\] is perpendicular to \[\overrightarrow{c}\] \[\Rightarrow \text{ }\,\,\left( \overrightarrow{a}+\overrightarrow{b} \right)\,.\,\overrightarrow{c}\text{ }=0\] \[=\,\,5({{b}_{1}}+1)+({{b}_{2}}+1)+\sqrt{2}\,(2\sqrt{2})\,\,=\,\,0\] \[=5{{b}_{1}}+{{b}_{2}}+10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\] solving (1) & (2) \[{{b}_{1}}=-3\text{ }and\text{ }{{b}_{2}}=5\] \[\Rightarrow \,\,\left| \overrightarrow{b} \right|\,\,=\,\,\sqrt{9+25+2}\,\,=\,\,6\]
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