A) \[2\sqrt{3}(\mathbf{i}-\mathbf{j}+\mathbf{k})\]
B) \[2\sqrt{3}(-\mathbf{i}+\mathbf{j}+\mathbf{k})\]
C) \[2\sqrt{3}(\mathbf{i}+\mathbf{j}-\mathbf{k})\]
D) \[2\sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})\]
Correct Answer: D
Solution :
Let \[l,m,n\] be the d.c's of \[\mathbf{r}.\] Then \[l=m=n\], (given) \ \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\Rightarrow 3{{l}^{2}}=1\Rightarrow l=\frac{1}{\sqrt{3}}=m=n\] Now, \[\mathbf{r}=|\mathbf{r}|(l\mathbf{i}+m\mathbf{j}+n\mathbf{k})=6\left( \frac{1}{\sqrt{3}}\mathbf{i}+\frac{1}{\sqrt{3}}\mathbf{j}+\frac{1}{\sqrt{3}}\mathbf{k} \right)\] Hence,\[\mathbf{r}=2\sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})\].
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