Vector Algebra Question 161
Question: 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]
Options:
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} $
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Answer:
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( \mathbf{i}-\frac{4}{3}\mathbf{j} ) $ , $ {\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}( \mathbf{i}-\frac{4}{3}\mathbf{j} )=2\mathbf{i}-\mathbf{j} $
or $ \mathbf{r}=\frac{1}{5}(4\mathbf{i}+3\mathbf{j})-\frac{6}{5}( \mathbf{i}-\frac{4}{3}\mathbf{j} )=-\frac{2}{5}\mathbf{i}+\frac{11}{5}\mathbf{j}. $
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