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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    Let \[\mathbf{b}=4\mathbf{i}+3\mathbf{j}\] and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively, are given by [IIT 1987]

    A) \[2\mathbf{i}-\mathbf{j},\,\,\frac{2}{5}\mathbf{i}+\frac{11}{5}\mathbf{j}\]

    B) \[2\mathbf{i}+\mathbf{j},\,\,-\frac{2}{5}\mathbf{i}+\frac{11}{5}\mathbf{j}\]

    C) \[2\mathbf{i}+\mathbf{j},\,-\frac{2}{5}\mathbf{i}-\frac{11}{5}\mathbf{j}\]

    D) \[2\mathbf{i}-\mathbf{j},\,\,-\frac{2}{5}\mathbf{i}+\frac{11}{5}\mathbf{j}\]

    Correct Answer: D

    Solution :

    • Let \[\mathbf{r}=\lambda \mathbf{b}+\mu \mathbf{c}\] and \[\mathbf{c}=\pm (x\mathbf{i}+y\mathbf{j}).\] Since \[\mathbf{c}\] and \[\mathbf{b}\] are perpendicular, we have \[4x+3y=0\]                   
    • \[\Rightarrow \mathbf{c}=\pm x\text{ }\left( \mathbf{i}-\frac{4}{3}\mathbf{j} \right)\],  \[\{\because \,y=-\frac{4}{3}x\}\]                   
    • Now, projection of \[\mathbf{r}\] on \[\mathbf{b}=\frac{\mathbf{r}\,.\,\mathbf{b}}{|\mathbf{b}|}=1\]                   
    • \[\Rightarrow \frac{(\lambda \mathbf{b}+\mu \mathbf{c})\,.\,\mathbf{b}}{|\mathbf{b}|}=\frac{\lambda \mathbf{b}\,.\,\mathbf{b}}{|\mathbf{b}|}=1\Rightarrow \lambda =\frac{1}{5}\]                   
    • Again, projection of \[\mathbf{r}\] on \[\mathbf{c}=\frac{\mathbf{r}\,.\,\mathbf{c}}{|\mathbf{c}|}=2\]                   
    • This gives \[\mu \,x=\frac{6}{5}\]\[\Rightarrow \mathbf{r}=\frac{1}{5}(4\mathbf{i}+3\mathbf{j})+\frac{6}{5}\left( \mathbf{i}-\frac{4}{3}\mathbf{j} \right)=2\mathbf{i}-\mathbf{j}\]                   
    • or \[\mathbf{r}=\frac{1}{5}(4\mathbf{i}+3\mathbf{j})-\frac{6}{5}\left( \mathbf{i}-\frac{4}{3}\mathbf{j} \right)=-\frac{2}{5}\mathbf{i}+\frac{11}{5}\mathbf{j}.\]


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