A plane electromagnetic wave is propagating along the direction $$\frac{\hat{i}+\hat{j}}{\sqrt{2}}$$, with its polarization along the direction $$\hat{k}$$. The correct form of the magnetic field of the wave would be (here $$B_0$$ is an appropriate constant):

For a monochromatic plane electromagnetic wave in free space we usually write the electric and magnetic fields in the form

$$\vec E = \vec E_0 \cos(\omega t - \vec k \cdot \vec r), \qquad \vec B = \vec B_0 \cos(\omega t - \vec k \cdot \vec r).$$

The three vectors $$\vec k,\; \vec E,\; \vec B$$ must satisfy two vector relations that follow directly from Maxwell’s equations:

1. All three are mutually perpendicular, so $$\vec k \cdot \vec E = 0$$ and $$\vec k \cdot \vec B = 0.$$

2. The magnetic field is obtained from the electric field through the vector product

$$\vec k \times \vec E = \dfrac{\omega}{c}\,\vec B,$$

where $$c$$ is the speed of light. The factor $$\dfrac{\omega}{c}$$ merely fixes the magnitudes, while the cross-product fixes the direction of $$\vec B$$ relative to $$\vec k$$ and $$\vec E$$. We therefore concentrate on directions; the overall constant will be written simply as $$B_0$$.

According to the statement of the problem, the wave is propagating along

$$\hat n = \dfrac{\hat i + \hat j}{\sqrt 2}.$$

Hence the wave-vector can be written as

$$\vec k = k\,\hat n = k\,\dfrac{\hat i + \hat j}{\sqrt 2},$$

where $$k = \dfrac{\omega}{c}$$ is the magnitude. The polarization (i.e. the direction of the electric field) is given to be along $$\hat k$$ (the positive $$z$$-axis):

$$\vec E_0 = E_0\,\hat k.$$

Because $$\vec k \cdot \vec E_0 = 0$$ is automatically satisfied (the dot-product of $$\dfrac{\hat i + \hat j}{\sqrt 2}$$ with $$\hat k$$ is zero), we proceed directly to the second relation. We compute the cross product $$\vec k \times \vec E_0$$ step by step:

First write the cross product explicitly:

$$\vec k \times \vec E_0 = \left(k\,\dfrac{\hat i + \hat j}{\sqrt 2}\right) \times \left(E_0\,\hat k\right) = kE_0\,\dfrac{1}{\sqrt 2}\;\bigl(\hat i + \hat j\bigr)\times\hat k.$$

Now evaluate each elementary cross product, recalling the right-handed basis rule $$\hat i \times \hat j = \hat k,\;\hat j \times \hat k = \hat i,\;\hat k \times \hat i = \hat j,$$ and therefore $$\hat i \times \hat k = -\hat j,\;\hat j \times \hat i = -\hat k,\;\hat k \times \hat j = -\hat i.$$ We need $$\hat i \times \hat k$$ and $$\hat j \times \hat k$$:

$$\hat i \times \hat k = -\hat j,\qquad \hat j \times \hat k = \hat i.$$

Substituting these results, we obtain

$$\bigl(\hat i + \hat j\bigr)\times\hat k = \hat i \times \hat k \;+\; \hat j \times \hat k = (-\hat j) + (\hat i) = \hat i - \hat j.$$

Therefore

$$\vec k \times \vec E_0 = kE_0\,\dfrac{1}{\sqrt 2}\;(\hat i - \hat j).$$

From the relation $$\vec k \times \vec E = \dfrac{\omega}{c}\,\vec B$$ we identify the direction of $$\vec B$$ to be exactly that of $$\hat i - \hat j$$ divided by $$\sqrt 2$$. All magnitude factors can be absorbed into a single constant $$B_0 = \dfrac{E_0}{c}$$, so we may write

$$\vec B_0 = B_0\,\dfrac{\hat i - \hat j}{\sqrt 2}.$$

The space-time dependence of the magnetic field must share the same argument $$\omega t - \vec k \cdot \vec r$$ as the electric field, because the wave propagates in the +$$\hat n$$ direction. Writing that explicitly we have

$$\vec B(\vec r,t) = B_0\,\dfrac{\hat i - \hat j}{\sqrt 2}\; \cos\!\Bigl(\omega t - \vec k \cdot \vec r\Bigr).$$

Using $$\vec k = k\,\dfrac{\hat i + \hat j}{\sqrt 2}$$, the scalar product $$\vec k \cdot \vec r$$ can be left indicated in vector form, exactly as it appears in the alternatives given. Hence the magnetic field is

$$\boxed{\; \vec B = B_0\,\dfrac{\hat i - \hat j}{\sqrt 2}\; \cos\!\left(\omega t - k\,\dfrac{\hat i + \hat j}{\sqrt 2}\cdot\vec r\right) \;} $$

This expression is identical to Option A when one recognises that the dot product with $$\vec r$$ is implicit. No other option has both the correct direction $$(\hat i - \hat j)/\sqrt 2$$ and the correct phase $$(\omega t - \vec k \cdot \vec r).$$

Hence, the correct answer is Option A.