Vector Algebra Question 315
Question: If $ \mathbf{r},.,\mathbf{i}=\mathbf{r},.,\mathbf{j}=\mathbf{r},.,\mathbf{k} $ and $ |\mathbf{r}|=3, $ then $ \mathbf{r}= $
Options:
A) $ \pm ,3,(\mathbf{i}+\mathbf{j}+\mathbf{k}) $
B) $ \pm ,\frac{1}{3},(\mathbf{i}+\mathbf{j}+\mathbf{k}) $
C) $ \pm ,\frac{1}{\sqrt{3}},(\mathbf{i}+\mathbf{j}+\mathbf{k}) $
D) $ \pm ,\sqrt{3},(\mathbf{i}+\mathbf{j}+\mathbf{k}) $
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Answer:
Correct Answer: D
Solution:
- Let $ \mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}. $ Since $ \mathbf{r}.\mathbf{i}=\mathbf{r}.\mathbf{j}=\mathbf{r}.\mathbf{k} $
$ \Rightarrow x=y=z $ …..(i) Also $ |\mathbf{r}|=\sqrt{x^{2}+y^{2}+z^{2}}=3\Rightarrow x=\pm \sqrt{3} $ , {By (i)} Hence the required vector $ \mathbf{r}=\pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k}). $ Trick: As the vector $ \pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k}) $ satisfies both the conditions.
BETA
