A) \[\hat{s}=\frac{4\hat{j}-3\hat{k}}{5}\]
B) \[\hat{s}=\frac{3\hat{j}-4\hat{j}}{5}\]
C) \[\hat{s}=\left( \frac{-3\hat{j}+4\hat{k}}{5} \right)\]
D) \[\hat{s}=\frac{-4\hat{j}+3\hat{j}}{5}\]
Correct Answer: C
Solution :
\[\vec{E}={{E}_{0}}\hat{n}\sin (\omega t+(6y-8z))\] \[={{E}_{0}}\hat{n}\sin (\omega t+\vec{k}.\vec{r})\]where \[\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\]and \[\hat{k}.\vec{r}=6y-8z\]\[\Rightarrow \vec{k}=6\hat{j}-8\hat{k}\] direction of propagation\[\hat{s}=-\hat{k}\]\[=\left( \frac{-3\hat{j}+4\hat{k}}{5} \right)\]
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