38. Equilateral $\angle A B C$ and equilateral $\triangle A B D$ are in two mutually perpendicular planes, then $\angle C A D=(\quad)$.
(A) $\cos ^{-1}\left(-\frac{1}{2}\right)$
(B) $\cos ^{-1}\left(\frac{1}{4}\right)$
(C) $\cos ^{-1}\left(-\frac{7}{16}\right)$
(D) $\frac{\pi}{2}$
(E) *
38. (B).

As shown in Figure 4, take the midpoint $E$ of $AB$. According to the problem, we know that $CE \perp AB$, $DE \perp AB$, $CE \perp DE$, and $CE = DE = \frac{\sqrt{3}}{2} AB$.
Thus, $CD = \sqrt{2} CE = \frac{\sqrt{6}}{2} AB$.
Then, $\cos \angle CAI$
$$
=\frac{2 \cdot AB^{2} - CD^{2}}{2 \cdot AB^{2}} = 1 - \frac{\left(\frac{\sqrt{6}}{2}\right)^{2}}{2} = \frac{1}{4}.
$$

Therefore, $\angle CAD = \cos^{-1}\left(\frac{1}{4}\right)$.