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  • Let <img alt="\vec{a}=\hat{i}+\hat{j}+\sqrt{2}\hat{k},\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+\sqrt{2}\hat{k}\And \vec{c}=5\hat{i}+\hat{j}+\sqrt{2}\hat{k}" src="https://entrancecor

Let \vec{a}=\hat{i}+\hat{j}+\sqrt{2}\hat{k},\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+\sqrt{2}\hat{k}\And \vec{c}=5\hat{i}+\hat{j}+\sqrt{2}\hat{k}   be three vectors such that the projection vector of \[\vec{b}\] on \[\vec{a}\] is \[\vec{a}\]. If \vec{a}+\vec{b} is perpendicular to \vec{c} ,then \left| {\vec{b}} \right|\ is equal to : 

Option: 1

\sqrt{22}


Option: 2

4


Option: 3

\sqrt{32}


Option: 4

6


Answers (1)

best_answer

Given, the projection vector of \vec{b}on \vec{a} is \vec{a}  ,

\frac{\vec{b}.\vec{a}}{\left| {\vec{a}} \right|}=\left| {\vec{a}} \right| \\ \vec{a}.\vec{b}={{\left| {\vec{a}} \right|}^{2}} \\ \vec{a}.\vec{b}={{b}_{1}}+{{b}_{2}}+2 \\ \left| {\vec{a}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{(\sqrt{2})}^{2}}} \\ \left| {\vec{a}} \right|=2 \\ {{b}_{1}}+{{b}_{2}}+2=4 \\ {{b}_{1}}+{{b}_{2}}=2 \\

If \vec{a}+\vec{b} is perpendicular to \vec{c},

(\vec{a}+\vec{b}).\vec{c}=0 \\ ((1+{{b}_{1}})\hat{i}+(1+{{b}_{2}})\hat{j}+2\sqrt{2}\hat{k}).(5\hat{i}+\hat{j}+\sqrt{2}\hat{k})=0 \\ 5(1+{{b}_{1}})+(1+{{b}_{2}})+4=0 \\ 10+5{{b}_{1}}+{{b}_{2}}=0 \\

Solving the equations,
{{b}_{1}}+{{b}_{2}}=2 \\ 5{{b}_{1}}+{{b}_{2}}=-10 \\

We get,
{{b}_{1}}=-3 \\ {{b}_{2}}=5 \\ \left| {\vec{b}} \right|=\sqrt{b_{1}^{2}+b_{2}^{2}+2} \\\\ \left| {\vec{b}} \right|=\sqrt{9+25+2} \\\\ \left| {\vec{b}} \right|=\sqrt{36} \\\\ \left| {\vec{b}} \right|=6 \\\\

 

Posted by

jitender.kumar

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