Let the $$x-z$$ plane be the boundary between two transparent media. Medium 1 in $$z \geq 0$$ has a refractive index of $$\sqrt{2}$$ and medium 2 with $$z < 0$$ has a refractive index of $$\sqrt{3}$$. A ray of light in medium 1 given by the vector $$\vec{A} = 6\sqrt{3}\,\hat{i} + 8\sqrt{3}\,\hat{j} - 10\hat{k}$$ is incident on the plane of separation. The angle of refraction in medium 2 is:

The interface is the $$x{-}y$$ plane $$z = 0$$, whose unit normal (drawn from medium 2 to medium 1) is $$\hat{n} = \hat{k}$$. The given incident ray in medium 1 is

$$\vec{A}=6\sqrt{3}\,\hat{i}+8\sqrt{3}\,\hat{j}-10\,\hat{k}$$

1. Angle of incidence
The angle $$\theta_i$$ between the ray and the normal is obtained from the dot product:

$$\cos\theta_i=\frac{\vec{A}\cdot\hat{n}}{|\vec{A}|}$$

$$\vec{A}\cdot\hat{n}= -10,\qquad |\vec{A}|=\sqrt{(6\sqrt{3})^{2}+(8\sqrt{3})^{2}+(-10)^{2}} =\sqrt{108+192+100}=20$$

Taking the magnitude (the sign of the dot product only tells the direction),

$$\cos\theta_i=\frac{10}{20}=\frac12 \;\;\Longrightarrow\;\; \theta_i = 60^\circ$$

2. Apply Snell’s law
Medium 1: $$n_1=\sqrt{2}$$, Medium 2: $$n_2=\sqrt{3}$$.
Snell’s law is $$n_1\sin\theta_i = n_2\sin\theta_r\;.$$

Insert the values:

$$\sqrt{2}\,\sin60^\circ=\sqrt{3}\,\sin\theta_r$$ $$\sqrt{2}\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}\,\sin\theta_r$$ $$\frac{\sqrt{6}}{2}=\sqrt{3}\,\sin\theta_r$$ $$\sin\theta_r=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{2}}{2}=\sin45^\circ$$

3. Angle of refraction
$$\theta_r = 45^\circ$$

Hence, the refracted ray makes an angle of $$45^\circ$$ with the normal.

Option A which is: $$45^\circ$$