• 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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