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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    Let \[\overset{\to }{\mathop{a}}\,=2\hat{i}-\hat{j}+k,\overset{\to }{\mathop{b}}\,=\hat{i}+2\hat{j}-\hat{k}\]and \[\overset{\to }{\mathop{c}}\,=\hat{i}+\hat{j}-2\overset{\to }{\mathop{k}}\,\]be three vectors. A vector of the type \[\overset{\to }{\mathop{b}}\,+\lambda \overset{\to }{\mathop{c}}\,\] for some scalar\[\lambda ,\] whose projection on \[\overset{\to }{\mathop{a}}\,\] is of magnitude\[\sqrt{\frac{2}{3}},\] is:                   JEE Main Online Paper (Held On 09 April 2013)

    A)                 \[2\hat{i}+\hat{j}+5\hat{k}\]                

    B)                 \[2\hat{i}+3\hat{j}-3\hat{k}\]                

    C)                 \[2\hat{i}-\hat{j}+5\hat{k}\]                

    D)                 \[2\hat{i}+3\hat{j}+3\hat{k}\]                

    Correct Answer: B

    Solution :

                   Given, \[a=2\hat{i}-\hat{j}+\hat{k},\,b=\hat{i}+2\hat{j}-\hat{k}\] and \[c=\hat{i}+\hat{j}-2\hat{k}\]                 Now, we have;  \[b+\lambda c=(1+\lambda )\hat{i}+(2+\lambda )\hat{j}+(-1-2\lambda )\hat{k}\] \[\therefore \] Projection of \[(b+\lambda c)\] on                 \[a=\left| \frac{(b+\lambda c)\cdot a}{|a|} \right|=\sqrt{\frac{2}{3}}\] (given) \[\Rightarrow \]               \[\left| \frac{2(1-\lambda )-(2+\lambda )\,+(-1-2\lambda )}{\sqrt{4+1+1}} \right|=\sqrt{\frac{2}{3}}\] \[\Rightarrow \]               \[\left| \frac{-\lambda -1}{\sqrt{6}} \right|=\frac{\sqrt{2}}{\sqrt{3}}\Rightarrow \,\,\lambda +1=2\Rightarrow \,\,\lambda =1\] \[\therefore \]  \[b+\lambda c=2\hat{i}+3\hat{j}-3\hat{k}\]                        


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