Finding directions of fast axis to cause circular polarisation

[Kaguro]

Homework Statement

Let $\vec E = (3 \hat i + 4 \hat j) exp[i(\omega t-kz)]$ represent an electromagnetic wave. Possible directions of the fast axis of a quarter wave plate which converts this wave into a circular wave are:


a) $\frac{1}{\sqrt{2}}[ 7 \hat i + \hat j]$ and $\frac{1}{\sqrt{2}}[ -\hat i + \hat j]$


b) $\frac{1}{\sqrt{2}}[ 3 \hat i -4 \hat j]$ and $\frac{1}{\sqrt{2}}[ 4\hat i -3 \hat j]$


c) $\frac{1}{\sqrt{2}}[ 3 \hat i -4 \hat j]$ and $\frac{1}{\sqrt{2}}[ 4\hat i + 3\hat j]$


a) $\frac{1}{\sqrt{2}}[ 7 \hat i - \hat j]$ and $\frac{1}{\sqrt{2}}[ \hat i + 7\hat j]$

Relevant Equations

None

The only thing I can think of is that to create a circularly polarized wave the axes of the quarter wave plate will have to be at 45 degrees to the E vector. Only then it can have both components on the slow and fast axis equal. Then the slow axis will cause delay and the resulting vector will rotate circularly. So the dot product of $\vec E$ with the axes should be equal.

The answer given is (c). But I don't know why.
Any guidance will be appreciated.

 

[anuttarasammyak]

45, 135, 225, 315 degree vector to $(3,4)$ are
$$(-1,7),(7,1),(1,-7),(-7,-1)$$
with normalization factor $\frac{1}{5\sqrt{2}}$ multiplied. I am afraid they are not in choice (a) to (d).

 

[Kaguro]

Just a misprint then.

Thank you for clearing my doubt.